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http://www.earlevel.com/main/2003/03/02/the-digital-state-variable-filter/ | 1,722,677,493,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00802.warc.gz | 43,186,425 | 18,387 | # The digital state variable filter
The digital state variable filter was described in Hal Chamberlin’s Musical Applications of Microprocessors. Derived by straight-forward replacement of components from the analog state variable fiter with digital counterparts, the digital state variable is a popular synthesizer filter, as was its analog counterpart.
The state variable filter has several advantages over biquads as a synthesizer filter. Lowpass, highpass, bandpass, and band reject are available simultaneously. Also, frequency and Q control are independent and their values calculated easily.
The frequency control coefficient, f, is defined as
where Fs is the sample rate and Fc is the filter’s corner frequency you want to set. The q coefficient is defined as
where Q normally ranges from 0.5 to inifinity (where the filter oscillates).
Like its analog counterpart, and biquads, the digital state variable has a cutoff slope of 12 dB/octave.
The main drawback of the digital state variable is that it becomes unstable at higher frequencies. It depends on the Q setting, but basically the upper bound of stability is about where f reaches 1, which is at one-sixth of the sample rate (8 kHz at 48 kHz). The only way around this is to oversample. A simple way to double the filter’s sample rate (and thereby double the filter’s frequency range) is to run the filter twice with the same input sample, and discard one output sample.
### As a sine oscillator
The state variable makes a great low frequency sine wave oscillator. Just set the Q to infinity, and make sure it has an impulse to get it started. Simply preset the delays (set the “cosine” delay to 1, or other peak value, and the other to 0) and run, and it will oscillate forever without instability, with fixed point or floating point. Even better, it gives you two waves in quadrature—simultaneous sine and cosine.
Simplified to remove unecessary parts, the oscillator looks like this:
For low frequencies, we can reduce the calculation of the f coefficient equation to
Here’s an example in C to show how easy this oscillator is to use; first initialize the oscillator amplitude, amp, to whatever amplitude you want (normally 1.0 for ±1.0 peak-to-peak output):
// initialize oscillator
sinZ = 0.0;
cosZ = amp;
Then, for every new sample, compute the sine and cosine components and use them as needed:
// iterate oscillator
sinZ = sinZ + f * cosZ;
cosZ = cosZ – f * sinZ;
The sine purity is excellent at low frequencies (becoming asymmetrical at high frequencies).
This entry was posted in Digital Audio, Filters, IIR Filters. Bookmark the permalink.
### 25 Responses to The digital state variable filter
1. TAL says:
Great article. Will try it. Like the idea of a filter that does not need complex volume and frequency corrction tables. Always used some type of biquad filters. Keep posting 🙂
2. Very nice post, Nigell! Could you please develop the 4th (and possible the 8th) order digital SVF. The following paper presents the development for the analog 4th SVF:
DENNIS A. BOHN, “A FOURTH-ORDER STATE VARIABLE FILTER FOR LINKWITZ-RILEY ACTIVE CROSSOVER DESIGNS” in AES Convention:74 (October 1983)
• Nigel Redmon says:
First, most higher-order filters are made by cascading lower-order filters—this is done to reduce the precision necessary for the higher-order poles. For instance, you can use two of these second-order SVFs in series for a fourth-order filter. The Moog-style digital filters are made from four first-order filters.
One key consideration in a synthesizer filter is how well it behaves when frequency (and perhaps resonance) are changed quickly; that’s why we don’t use direct form biquads as synthesizer filters. I hope to find time to write an article on an improved SVF suitable for synthesizers for an upcoming post.
• John says:
I’m designing a synthesizer with a biquad that accepts a 1/q value (I considered hardwiring it as a Butterworth) and can have its frequency driven by an LFO. The output of either phase modulation or an alias-suppressed trivial waveform generator is the input. That’s bad?
By the way, testing this for 60 seconds at 96,000Hz fs with a 300Hz f0 and an amplitude of 1 gives me a maximum sine value of 1.0000481949464535 and appears to be stable. That little extra tacked on the end is a tiny amount.
Cascading two 2nd order state-variable filters (SVF) is indeed straightforward. The result, however, does not feature the expected 4th order LP/HP/BP/BR simultaneous outputs — rather, only one of the outputs is as expected.
Without loss of generality, let’s assume that the LP output of the SVF I is connected to the SVF II input. In such case, it is true that one has at the LP output of SVF II the expected 4th order lowpass filter, however the remaining SVF II outputs are LP-XX cascades of little meaning or use. The same applies to the HP, BP, and BR.
Being mostly focuseded on implementing a time-invariant 4th order crossover using a single 4th order SVF, I look after topology in which the complementary 4th order low and highpass outputs are simultaneously available.
• Nigel Redmon says:
OK, I see now that your goal is a crossover and not a lowpass synth filter. I suggest focusing your search on “Linkwitz-Riley” in particular, rather than fourth-order state variable.
4. Q says:
Isn’t there a delay in the output back to input line missing?
• Nigel Redmon says:
There just needs to be a delay somewhere in the loop, as you can see in the diagram. I’ll try to present a more modern, improved version in the future.
5. ... says:
“The main drawback of the digital state variable is that it becomes unstable at higher frequencies.”
Is this an advantage over biquads for low frequency filters? Biquads work fine at high frequencies, but are supposed to be implemented in extra precision at very low frequencies because they have numerical problems, while SVF need special attention at high frequencies but work well at low frequencies?
• Nigel Redmon says:
Numerically, this state variable filter is very good at low frequencies (the direct form biquads aren’t). There are ways to improve the Chamberlin state variable. The most obvious is oversampling, with a cheap and dirty version being to simply run the filter twice (with the same input) and discard the first result. Andrew Simper shows how to use trapezoidal integration for much-improved version of the state variable filter: http://www.cytomic.com/technical-papers
6. Josh Marler says:
Hi Nigel,
I’ve been a huge fan of your site whilst on my journey to develop some digital synth’s and you’ve been kind enough to even help me with a question or two in the past.
I was wondering if you had any plans to do an article on the implementation of a working SVF ?
I’ve managed to build up a fairly decent understand of IIR filters and biquads etc but want to implement a filter that responds well to parameter changes and modulated cuttoff controls etc.
Your WaveTable oscillator class was an absolute gold mine and gem of a read for a dsp newbie like myself and gave me hope again whilst implementing my own synth!
Is anything similar in the pipeline for an svf ?
Failing that would you or anyone else reading happen to know of any resources that provide a well explained example of implementing a SVF in c++ and caculating the various coefficients etc ?
Huge thanks again for sharing all your know how.
Josh
7. mclaren says:
This article represents an outstanding introduction to a very deep topic. In reality, these SVFs are actually digital waveguides, which can be used to generate waveforms of artbitrary complexity. Carried to the extreme, these kinds of digital waveguides lead to physical modeling, a system able to reproduce acoustic instruments with eerie accuracy (as well as able to model analog circuits, with some slight changes).
I hope you plan to go into a lot more depth on this subject, because it’s of tremendous importance now that CPUs and DSP chips have gotten fast enough to run IIR filter code in real time. Synths like the Nord modular or the Yamaha VL-1 use this kind of code to produce amazing sounds.
See Julius O. Smith III’s article “Digital Waveguide Architectures for Virtual Instruments” for a lot more detail.
8. kavi says:
I am student from INDIA.i want to know what is gain constant,pole frequency and pole selectivity generally..and for state variable filter how it will derived from transfer function? i am weak in this area..can u clarify it??
9. Dom Smat says:
I just wanted to stop by and say thank you for this really informative article. It’s most helpful!
10. Ian Benton says:
Something doesn’t look quite right. . .
An op-amp state variable consists of two cascaded identical integrators (one which produces BP output followed by one which produces LP output) with an adder to close the loop. The BP output goes into the input to the second integrator. In your digital implementation the input to the second integrator comes from the output of the delay, not the BP output. Would that make a difference?
• Nigel Redmon says:
Right. The reason is that it would be a delay-free loop (the output of the first thing depending on the output of the second thing that depends on the output of the first…). There are better ways to go, at the expense of a little more computation, such as Andrew Simper’s version with trapezoidal integrators (search for andrew simper state variable). Also, you might come across “zero-delay feedback” filters, which aren’t really a delay-free loop, but try to approximate one with more computation.
• Ian Benton says:
I also found your explanation in Hal Chamberlain’s book.
I was expecting BOTH delays in the loop, as it seemed more like the two cascaded integrators in the analogue version.
11. Kim says:
Great article.
You write that the sine oscillator “will oscillate forever without instability, with fixed point or floating point.”
Can you prove the sine oscillator is guaranteed stable no matter the precision used?
How can you be sure it will always be stable?
The background for asking this is that I started out implemented the sliding Goertzel algorithm using IIR direct form 2 for the resonator. For sliding Goertzel, the poles are always on the unit circle and the filter is marginally stable. Even when using double precision floating point, the output slowly drifts. The problem is a limited precision accumulator.
One solution to this problem is to move the poles inside the unit circle (and move the cancelling zeros). In that case, I would have to figure out how much the poles had to be moved for the direct form 2 realization to become stable. Of course, I could always just move the poles longer than needed to be on the safe side but it would be a lot better to know the exact requirement. Any hints/links to relevant articles would be appreciated.
Then the sliding goertzel was implemented with the resonator in state variable form, like described in this article. When the resonator is realized by using the state variable structure, the bandpass output is used and q is zero (infinite Q), making it similar to the sine oscillator. In this case, the resonator seems to be stable with both single and double precision floating point. However, I do not know if it really is stable or I just have not tested with an input that provokes the instability. If I could prove that the output is guaranteed stable, I can keep Q at infinity, otherwise I will have to find a suitable Q that guarantees stability for a given precision.
Finally I have not considered floating point denormals and how it affects performance but stability issues come first.
• Nigel Redmon says:
I have only a superficial familiarity with sliding Geortzel, having seen articles and discussion, but had no need. I can’t point you to anything you haven’t already web-searched for (this seems like a nice overview of issues, by Eric Jacobsen and Richard Lyons: The Sliding DFT). The comp.dsp group would be a good place to ask.
• Kim says:
Thanks Nigel. I will try that.
Could you elaborate on how you can be certain that the sine oscillator mentioned in your article will be stable regardless of using floating or fixed point?
• Nigel Redmon says:
Chamberlin asserts the amplitude stability with integers in his book, though from observation (he tested with 8-bit arithmetic). I believe I read later that it was true of floating point as well. I’ve never tried to prove it.
12. David says:
Hello Nigel,
great article, thank you very much.
One thing I’d like to know is, if I can plot the SVF somehow like you did it with the biquad filter and how would yo do that?
Thank you
David
• Nigel Redmon says:
The biquad calculator is done by calculating the response around the unit circle based on pole and zero positions dictated by the biquad coefficients. While the SVF is fundamentally equivalent, I can’t tell you an easy way to modify what I have for that, offhand. But plotting the response of and arbitrary system is pretty easy: Start with a system initially internally to zero (or run a string of zeros through it—not necessary for an SVF with the delays zeroed), then a “1” (anything non-zero, especially if your code is floating point), followed by more zero samples. Then take an FFT of the output and plot it. Longer FFTs will have better frequency resolution, and there’s the caveat that lower filter features (a lowpass set to a low cutoff frequency) naturally have a longer decay time, but as long as the FFT isn’t too short (poor plotting resolution anyway), you can get a pretty good picture without bothering to window. To check the SVF, pick an FFT size that fits the number of horizontal pixels on your graph (2048 or 4096, say) and that should give you decent results.
• David says:
Thank you! That really helped me. | 3,041 | 13,856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.926981 |
http://www.fitnessforweightloss.com/how-many-calories-do-i-burn-playing-volleyball/ | 1,516,383,795,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888077.41/warc/CC-MAIN-20180119164824-20180119184824-00222.warc.gz | 448,471,327 | 9,745 | # How many calories do I burn playing volleyball?
## You can burn 105 – 765 calories playing 30 minutes of volleyball.
It depends on how much you weigh, how much you move, and if you’re playing for fun or competitively. Below is an estimate of how many calories you can burn playing volleyball for 30 minutes.
### Non-competitive volleyball (6-9 people per team)
• If you weigh 150 lbs, you can burn approximately 105 calories in 30 minutes.
• If you weigh 200 lbs, you can burn approximately 145 calories in 30 minutes.
• If you weigh 250 lbs, you can burn approximately 180 calories in 30 minutes.
• If you weigh 300 lbs, you can burn approximately 215 calories in 30 minutes.
• If you weigh 350 lbs, you can burn approximately 250 calories in 30 minutes.
• If you weigh 400 lbs, you can burn approximately 285 calories in 30 minutes.
### Competitive volleyball
• If you weigh 150 lbs, you can burn approximately 285 calories in 30 minutes.
• If you weigh 200 lbs, you can burn approximately 380 calories in 30 minutes
• If you weigh 250 lbs, you can burn approximately 475 calories in 30 minutes.
• If you weigh 300 lbs, you can burn approximately 575 calories in 30 minutes.
• If you weigh 350 lbs, you can burn approximately 670 calories in 30 minutes.
• If you weigh 400 lbs, you can burn approximately 765 calories in 30 minutes.
Sources:
Ainsworth BE, Haskell WL, Whitt MC, et al. Compendium of physical activities: an update of activity codes and MET intensities. Med. Sci. Sports Exerc., Vol. 32, No. 9, Suppl., pp. S498–S516, 2000.
Calorie calculations from equation: (METs x 3.5 x body weight in kg)/200 = calories/minute | 419 | 1,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-05 | latest | en | 0.888825 |
http://www.physicsforums.com/showthread.php?s=442a1a28e28869d88f6164a09dd72511&p=4593598 | 1,397,634,002,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00459-ip-10-147-4-33.ec2.internal.warc.gz | 615,329,394 | 7,325 | # Very basic linear algebra question
by 1MileCrash
Tags: algebra, basic, linear
P: 1,226 So as I'm preparing for finals, I'm wondering: The multiplication of two matrices is only defined under special circumstances regarding the dimensions of the matrices. Doesn't that require that compositions of linear transformations are only defined in the same circumstances? I can't imagine not being able to not define a composition of linear transformations, can someone demonstrate this?
P: 205 If $f:V\to W$ and $g:X\to Y$ are linear transformations, it only makes sense to talk about the composition $g\circ f$ if $W=X$. In particular, if $f:\mathbb R^K \to \mathbb R^L$ and $g:\mathbb R^J \to \mathbb R^I$ are linear transformations, it only makes sense to talk about the composition $g\circ f$ if $\mathbb R^L=\mathbb R^J$, i.e. if $L=J.$ Phrasing the last point a different way now: If $F$ is an $L\times K$ matrix and $G$ is an $I\times J$ matrix, it only makes sense to talk about the matrix product $GF$ if $L=J$. So the matrix dimension rule you learned is really there exactly because only certain functions can be composed. The expression $g\circ f$ only has meaning if the outputs of $f$ are valid inputs for $g$.
Mentor
P: 20,933
Quote by 1MileCrash So as I'm preparing for finals, I'm wondering: The multiplication of two matrices is only defined under special circumstances regarding the dimensions of the matrices. Doesn't that require that compositions of linear transformations are only defined in the same circumstances? I can't imagine not being able to not define a composition of linear transformations, can someone demonstrate this?
If A is an m X n matrix, and B is an n X p matrix, then the product AB is defined, and will be an m X p matrix.
A linear transformation TA: Rn → Rm takes vectors from Rn and maps them to vectors in Rm. A matrix for TA will by m X n. Think about how TB would have to be defined (in terms of its domain and codomain) so that the composition TA ° TB would make sense. It might be helpful to use constants for the dimensions.
P: 1,226
## Very basic linear algebra question
Crystal clear, thanks you two.
Related Discussions Calculus & Beyond Homework 3 Linear & Abstract Algebra 6 Calculus & Beyond Homework 3 Calculus & Beyond Homework 2 Calculus & Beyond Homework 16 | 577 | 2,324 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2014-15 | latest | en | 0.89998 |
http://de.metamath.org/mpegif/cdleme7aa.html | 1,611,767,376,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704828358.86/warc/CC-MAIN-20210127152334-20210127182334-00337.warc.gz | 25,614,012 | 10,828 | Mathbox for Norm Megill < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > cdleme7aa Structured version Visualization version Unicode version
Theorem cdleme7aa 33820
Description: Part of proof of Lemma E in [Crawley] p. 113. Lemma leading to cdleme7ga 33826 and cdleme7 33827. (Contributed by NM, 7-Jun-2012.)
Hypotheses
Ref Expression
cdleme4.l
cdleme4.j
cdleme4.m
cdleme4.a
cdleme4.h
cdleme4.u
cdleme4.f
cdleme4.g
Assertion
Ref Expression
cdleme7aa
Proof of Theorem cdleme7aa
StepHypRef Expression
1 simp33 1047 . 2
2 simp11l 1120 . . . 4
3 simp2ll 1076 . . . 4
4 simp2rl 1078 . . . 4
5 simp11 1039 . . . . 5
6 simp12 1040 . . . . 5
7 simp13 1041 . . . . 5
8 simp31 1045 . . . . 5
9 cdleme4.l . . . . . 6
10 cdleme4.j . . . . . 6
11 cdleme4.m . . . . . 6
12 cdleme4.a . . . . . 6
13 cdleme4.h . . . . . 6
14 cdleme4.u . . . . . 6
159, 10, 11, 12, 13, 14lhpat2 33622 . . . . 5
165, 6, 7, 8, 15syl112anc 1273 . . . 4
17 hllat 32941 . . . . . . . 8
182, 17syl 17 . . . . . . 7
19 simp12l 1122 . . . . . . . 8
20 eqid 2453 . . . . . . . . 9
2120, 10, 12hlatjcl 32944 . . . . . . . 8
222, 19, 7, 21syl3anc 1269 . . . . . . 7
23 simp11r 1121 . . . . . . . 8
2420, 13lhpbase 33575 . . . . . . . 8
2523, 24syl 17 . . . . . . 7
2620, 9, 11latmle2 16335 . . . . . . 7
2718, 22, 25, 26syl3anc 1269 . . . . . 6
2814, 27syl5eqbr 4439 . . . . 5
29 simp2lr 1077 . . . . 5
30 nbrne2 4424 . . . . . 6
3130necomd 2681 . . . . 5
3228, 29, 31syl2anc 667 . . . 4
339, 10, 12hlatexch1 32972 . . . 4
342, 3, 4, 16, 32, 33syl131anc 1282 . . 3
35 simp2l 1035 . . . . . 6
36 simp32 1046 . . . . . 6
379, 10, 11, 12, 13, 14cdleme4 33816 . . . . . 6
385, 19, 7, 35, 36, 37syl131anc 1282 . . . . 5
3910, 12hlatjcom 32945 . . . . . 6
402, 3, 16, 39syl3anc 1269 . . . . 5
4138, 40eqtrd 2487 . . . 4
4241breq2d 4417 . . 3
4334, 42sylibrd 238 . 2
441, 43mtod 181 1
Colors of variables: wff setvar class Syntax hints: wn 3 wi 4 wa 371 w3a 986 wceq 1446 wcel 1889 wne 2624 class class class wbr 4405 cfv 5585 (class class class)co 6295 cbs 15133 cple 15209 cjn 16201 cmee 16202 clat 16303 catm 32841 chlt 32928 clh 33561 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1671 ax-4 1684 ax-5 1760 ax-6 1807 ax-7 1853 ax-8 1891 ax-9 1898 ax-10 1917 ax-11 1922 ax-12 1935 ax-13 2093 ax-ext 2433 ax-rep 4518 ax-sep 4528 ax-nul 4537 ax-pow 4584 ax-pr 4642 ax-un 6588 This theorem depends on definitions: df-bi 189 df-or 372 df-an 373 df-3an 988 df-tru 1449 df-ex 1666 df-nf 1670 df-sb 1800 df-eu 2305 df-mo 2306 df-clab 2440 df-cleq 2446 df-clel 2449 df-nfc 2583 df-ne 2626 df-ral 2744 df-rex 2745 df-reu 2746 df-rab 2748 df-v 3049 df-sbc 3270 df-csb 3366 df-dif 3409 df-un 3411 df-in 3413 df-ss 3420 df-nul 3734 df-if 3884 df-pw 3955 df-sn 3971 df-pr 3973 df-op 3977 df-uni 4202 df-iun 4283 df-iin 4284 df-br 4406 df-opab 4465 df-mpt 4466 df-id 4752 df-xp 4843 df-rel 4844 df-cnv 4845 df-co 4846 df-dm 4847 df-rn 4848 df-res 4849 df-ima 4850 df-iota 5549 df-fun 5587 df-fn 5588 df-f 5589 df-f1 5590 df-fo 5591 df-f1o 5592 df-fv 5593 df-riota 6257 df-ov 6298 df-oprab 6299 df-mpt2 6300 df-1st 6798 df-2nd 6799 df-preset 16185 df-poset 16203 df-plt 16216 df-lub 16232 df-glb 16233 df-join 16234 df-meet 16235 df-p0 16297 df-p1 16298 df-lat 16304 df-clat 16366 df-oposet 32754 df-ol 32756 df-oml 32757 df-covers 32844 df-ats 32845 df-atl 32876 df-cvlat 32900 df-hlat 32929 df-psubsp 33080 df-pmap 33081 df-padd 33373 df-lhyp 33565 This theorem is referenced by: cdleme7c 33823
Copyright terms: Public domain W3C validator | 1,991 | 3,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-04 | latest | en | 0.059517 |
https://rehabilitationrobotic.com/what-is-the-square-of-1-16/ | 1,653,370,613,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662564830.55/warc/CC-MAIN-20220524045003-20220524075003-00569.warc.gz | 432,520,398 | 9,673 | Close
1/4
## What’s the square root of 1 16th?
sqrt(1/16) = 14 = 0.25.
## What is .0625 as a fraction?
Now you know that . 625 as a fraction is 5/8.
12.5
## What is 25% as a whole number?
Example
Problem Write 25% as a simplified fraction and as a decimal.
Simplify the fraction by dividing the numerator and denominator by the common factor 25.
Write as a decimal. 25% = = 0.25 You can also just move the decimal point in the whole number 25 two places to the left to get 0.25.
## What is a 8th of something?
One eighth is one part of eight equal sections. Two eighths is one quarter and four eighths is a half. It’s easy to split an object, like a cake, into eighths if you make them into quarters and then divide each quarter in half.
## Can you simplify 5 8?
58 is already in the simplest form. It can be written as 0.625 in decimal form (rounded to 6 decimal places).
## What is 2/8 in the lowest term?
Steps to simplifying fractions Therefore, 2/8 simplified to lowest terms is 1/4.
## What is the Simplify of 4 8?
Steps to simplifying fractions Therefore, 4/8 simplified to lowest terms is 1/2.
## What is the fraction 4/8 equal to?
Decimal and Fraction Conversion
Fraction Equivalent Fractions
5/8 10/16 20/32
7/8 14/16 28/32
1/9 2/18 4/36
2/9 4/18 8/36
1.3333
## How do you write 4 divided by 3?
4 divided by 3 is equal to 1 with a remainder of 1 (4 / 3 = 1 R. 1). When you divide 4 by 3, you are not left with an equal number of groups because 3…
## How do you do 4 divided by 9?
4 divided by 9 is equal to the fraction 4/9, or the repeating decimal 0.44444…, where the 4s go on forever past the decimal point.
## What is 4/9 in a number?
4/9 as a decimal is 0.44444444444444.
54
## What month are intelligent babies born?
Those born in September are, apparently, the smartest out of the entire year. According to Marie Claire, a study published in the National Bureau of Economic Research found that there’s a clear correlation between the month during which you were born and how smart you are.
## Which is the best month?
Top 10 Best Months of The Year
1. December. The perfect month for kids here’s why.
2. July. 1: My favorite month is July for a few reasons.
3. June. June is actually when my birthday is!
4. October. I know, nobody likes october because it isn’t December or in the summer.
5. May.
6. August.
7. April.
8. March.
## What is the saddest month?
Why January is Known As the Most Depressing Month of the Year.
May
spring
2021-06-10 | 725 | 2,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-21 | latest | en | 0.945762 |
https://www.mrexcel.com/board/threads/index-match-when-the-target-column-is-not-known.1094354/ | 1,638,678,641,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363135.71/warc/CC-MAIN-20211205035505-20211205065505-00280.warc.gz | 976,296,649 | 18,736 | # INDEX MATCH when the target column is not known
#### tonkerthomas
##### Board Regular
I've tried to find another thread that deals with this topic, without success, so here goes:
I have a very large block of data: thousands of rows by hundreds of columns. I need to identify a known value in that block and return the value two cells to the right of it. I can use INDEX to ascertain which row of my data the value will be in, but it could be in any column. How do I locate the value, and how do I offset my return?
Example:
<tbody>
</tbody>
I'm looking to find Clive, say, and want to return Bert. I know that Clive is in team B but I don't know where in the columns his name appears so I can't INDEX MATCH with Colhead2. I'm sure there's a way of doing this but I don't know how and an hour of Googling hasn't helped me. So, can you guys?
Jeff
Last edited:
### Excel Facts
How to calculate loan payments in Excel?
Use the PMT function: =PMT(5%/12,60,-25000) is for a \$25,000 loan, 5% annual interest, 60 month loan.
#### Fluff
##### MrExcel MVP, Moderator
+Fluff New.xlsm
ABCDEFGHIJK
3TeamBEricCliveAgnesBertDoraRitaCliveBert
4TeamCKatieMaryJakeWillKikiHarry
Lookup
Cell Formulas
RangeFormula
K3K3=INDEX(B2:G4,MATCH(J2,A2:A4,0),MATCH(J3,INDEX(B2:G4,MATCH(J2,A2:A4,0),0),0)+2)
Last edited:
#### tonkerthomas
##### Board Regular
Aha! An INDEX MATCH INDEX MATCH! That's logical, but sadly, too complicated for this bear of very little brain to figure out by himself. Thanks a million, Fluff - your time is very much appreciated.
Jeff
#### Fluff
##### MrExcel MVP, Moderator
You're welcome & thanks for the feedback
#### donblue
##### New Member
try this =INDEX(\$A:\$A,(ROW()-1)*3+COLUMN(A1)).
I used it to make print out available in 3 columns similar to Words print 'number of columns' excel2013
#### Fluff
##### MrExcel MVP, Moderator
Did you realise that this thread is over a year old?
Also you formula does not do what the OP was asking for.
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Go back | 844 | 3,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-49 | latest | en | 0.90225 |
https://pastebin.com/cKddCUrD | 1,723,363,535,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640983659.65/warc/CC-MAIN-20240811075334-20240811105334-00036.warc.gz | 358,102,404 | 6,583 | # 224
Jul 2nd, 2021 (edited)
166
0
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1. class Solution {
2. public:
3. int calculate(string s) {
4. long long int result=0;
5. long long int num=0, sign=1;
6. stack<long long int> nums, ops;
7.
8. for(int i=0; i<s.size(); ++i){
9. char c=s[i];
10. if(c>='0' && c<='9'){
11. num=10*num + c-'0'; /// For case: "23"
12. }else if(c=='+'){
13. result += num*sign; /// everytime meets an operator, sum up previous number
14. num=0;
15. sign=1; /// sign is the sign of next number
16. }else if(c=='-'){
17. result += num*sign; /// everytime meets an operator, sum up previous number
18. num=0;
19. sign=-1;
20. }else if(c=='('){
21. nums.push(result); /// ...(1+3+(..xx..)+...)... before go into cur (..xx..), record the forefront result (in this case it is 1+3 ) into nums array
22. ops.push(sign); // record cur (..xx..) sign
23. result=0; // result is to record the total value in the cur (..xx..)
24. sign=1;
25. }else if(c==')' && ops.size()){
26. result += num*sign; /// For case: 1-(5)
27. num=0;
28. result = result*ops.top() + nums.top(); // ...(1+3+(..xx..)+...)... sum up cur (..xx..) and the forefront result (in this case it is 1+3 )
29. nums.pop();
30. ops.pop();
31. }
32. }
33. result += num*sign; /// For the last one operation. consider the case: 1+2+3
34. return result;
35. }
36. };
37.
38. -----------------------------
39.
40. class Solution {
41. public:
42. int calculate(string s) {
43. int n = s.size(), ret = 0, sign = 1;
44. stack<int> stk;
45.
46. for (int i = 0; i < n; i++) {
47. if (isdigit(s[i])) {
48. int num = s[i] - '0';
49.
50. while (i + 1 < n && isdigit(s[i + 1])) {
51. num = num * 10 + (s[i + 1] - '0');
52. i++;
53. }
54.
55. ret += sign * num;
56. } else if (s[i] == '+') {
57. sign = 1;
58. } else if (s[i] == '-') {
59. sign = -1;
60. } else if (s[i] == '(') {
61. stk.push(ret);
62. stk.push(sign);
63. ret = 0;
64. sign = 1;
65. } else if (s[i] == ')') {
66. ret *= stk.top();
67. stk.pop();
68. ret += stk.top();
69. stk.pop();
70. }
71. }
72.
73. return ret;
74. }
75. }; | 827 | 2,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-33 | latest | en | 0.091381 |
https://www.topperlearning.com/answer/if-sin-1x-sin-1y-sin-1z-pi-prove-that-i-x-1-x-2-1-2-y-1-y-2-1-2-z-1-z-2-1-2-2xyz-ii-x-4-y-4-z-4-4x-2-y-2-z-2-2-x-2-y-2-y-2-z-2-z-2-x-2/daor0w22 | 1,620,570,403,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988986.98/warc/CC-MAIN-20210509122756-20210509152756-00072.warc.gz | 1,093,504,170 | 44,681 | # If sin^-1x+sin^-1y+sin^-1z=pi,prove that (i) x(1-x^2)^1/2 +y(1-y^2)^1/2 +z(1-z^2)^1/2=2xyz(ii) x^4 +y^4 +z^4 +4x^2 y^2 z^2 =2(x^2 y^2 +y^2 z^2+z^2-- x^2
## Concept Videos
### STUDY RESOURCES
REGISTERED OFFICE : First Floor, Empire Complex, 414 Senapati Bapat Marg, Lower Parel, Mumbai - 400013, Maharashtra India. | 150 | 318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-21 | latest | en | 0.401442 |
https://www.nag.com/numeric/nl/nagdoc_26.1/nagdoc_fl26.1/html/f01/f01eff.html | 1,632,554,418,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057598.98/warc/CC-MAIN-20210925052020-20210925082020-00275.warc.gz | 907,205,845 | 6,740 | # NAG Library Routine Document
## 1Purpose
f01eff computes the matrix function, $f\left(A\right)$, of a real symmetric $n$ by $n$ matrix $A$. $f\left(A\right)$ must also be a real symmetric matrix.
## 2Specification
Fortran Interface
Subroutine f01eff ( uplo, n, a, lda, f,
Integer, Intent (In) :: n, lda Integer, Intent (Inout) :: iuser(*), ifail Integer, Intent (Out) :: iflag Real (Kind=nag_wp), Intent (Inout) :: a(lda,*), ruser(*) Character (1), Intent (In) :: uplo External :: f
#include nagmk26.h
void f01eff_ (const char *uplo, const Integer *n, double a[], const Integer *lda, void (NAG_CALL *f)(Integer *iflag, const Integer *n, const double x[], double fx[], Integer iuser[], double ruser[]),Integer iuser[], double ruser[], Integer *iflag, Integer *ifail, const Charlen length_uplo)
## 3Description
$f\left(A\right)$ is computed using a spectral factorization of $A$
$A = Q D QT ,$
where $D$ is the diagonal matrix whose diagonal elements, ${d}_{i}$, are the eigenvalues of $A$, and $Q$ is an orthogonal matrix whose columns are the eigenvectors of $A$. $f\left(A\right)$ is then given by
$fA = Q fD QT ,$
where $f\left(D\right)$ is the diagonal matrix whose $i$th diagonal element is $f\left({d}_{i}\right)$. See for example Section 4.5 of Higham (2008). $f\left({d}_{i}\right)$ is assumed to be real.
## 4References
Higham N J (2008) Functions of Matrices: Theory and Computation SIAM, Philadelphia, PA, USA
## 5Arguments
1: $\mathbf{uplo}$ – Character(1)Input
On entry: if ${\mathbf{uplo}}=\text{'U'}$, the upper triangle of the matrix $A$ is stored.
If ${\mathbf{uplo}}=\text{'L'}$, the lower triangle of the matrix $A$ is stored.
Constraint: ${\mathbf{uplo}}=\text{'U'}$ or $\text{'L'}$.
2: $\mathbf{n}$ – IntegerInput
On entry: $n$, the order of the matrix $A$.
Constraint: ${\mathbf{n}}\ge 0$.
3: $\mathbf{a}\left({\mathbf{lda}},*\right)$ – Real (Kind=nag_wp) arrayInput/Output
Note: the second dimension of the array a must be at least ${\mathbf{n}}$.
On entry: the $n$ by $n$ symmetric matrix $A$.
• If ${\mathbf{uplo}}=\text{'U'}$, the upper triangular part of $A$ must be stored and the elements of the array below the diagonal are not referenced.
• If ${\mathbf{uplo}}=\text{'L'}$, the lower triangular part of $A$ must be stored and the elements of the array above the diagonal are not referenced.
On exit: if ${\mathbf{ifail}}={\mathbf{0}}$, the upper or lower triangular part of the $n$ by $n$ matrix function, $f\left(A\right)$.
4: $\mathbf{lda}$ – IntegerInput
On entry: the first dimension of the array a as declared in the (sub)program from which f01eff is called.
Constraint: ${\mathbf{lda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
5: $\mathbf{f}$ – Subroutine, supplied by the user.External Procedure
The subroutine f evaluates $f\left({z}_{i}\right)$ at a number of points ${z}_{i}$.
The specification of f is:
Fortran Interface
Subroutine f ( n, x, fx,
Integer, Intent (In) :: n Integer, Intent (Inout) :: iflag, iuser(*) Real (Kind=nag_wp), Intent (In) :: x(n) Real (Kind=nag_wp), Intent (Inout) :: ruser(*) Real (Kind=nag_wp), Intent (Out) :: fx(n)
#include nagmk26.h
void f (Integer *iflag, const Integer *n, const double x[], double fx[], Integer iuser[], double ruser[])
1: $\mathbf{iflag}$ – IntegerInput/Output
On entry: iflag will be zero.
On exit: iflag should either be unchanged from its entry value of zero, or may be set nonzero to indicate that there is a problem in evaluating the function $f\left(x\right)$; for instance $f\left(x\right)$ may not be defined, or may be complex. If iflag is returned as nonzero then f01eff will terminate the computation, with ${\mathbf{ifail}}=-{\mathbf{6}}$.
2: $\mathbf{n}$ – IntegerInput
On entry: $n$, the number of function values required.
3: $\mathbf{x}\left({\mathbf{n}}\right)$ – Real (Kind=nag_wp) arrayInput
On entry: the $n$ points ${x}_{1},{x}_{2},\dots ,{x}_{n}$ at which the function $f$ is to be evaluated.
4: $\mathbf{fx}\left({\mathbf{n}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: the $n$ function values. ${\mathbf{fx}}\left(\mathit{i}\right)$ should return the value $f\left({x}_{\mathit{i}}\right)$, for $\mathit{i}=1,2,\dots ,n$.
5: $\mathbf{iuser}\left(*\right)$ – Integer arrayUser Workspace
6: $\mathbf{ruser}\left(*\right)$ – Real (Kind=nag_wp) arrayUser Workspace
f is called with the arguments iuser and ruser as supplied to f01eff. You should use the arrays iuser and ruser to supply information to f.
f must either be a module subprogram USEd by, or declared as EXTERNAL in, the (sub)program from which f01eff is called. Arguments denoted as Input must not be changed by this procedure.
Note: f should not return floating-point NaN (Not a Number) or infinity values, since these are not handled by f01eff. If your code inadvertently does return any NaNs or infinities, f01eff is likely to produce unexpected results.
6: $\mathbf{iuser}\left(*\right)$ – Integer arrayUser Workspace
7: $\mathbf{ruser}\left(*\right)$ – Real (Kind=nag_wp) arrayUser Workspace
iuser and ruser are not used by f01eff, but are passed directly to f and may be used to pass information to this routine.
8: $\mathbf{iflag}$ – IntegerOutput
On exit: ${\mathbf{iflag}}=0$, unless you have set iflag nonzero inside f, in which case iflag will be the value you set and ifail will be set to ${\mathbf{ifail}}=-{\mathbf{6}}$.
9: $\mathbf{ifail}$ – IntegerInput/Output
On entry: ifail must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this argument you should refer to Section 3.4 in How to Use the NAG Library and its Documentation for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this argument, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6Error Indicators and Warnings
If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
${\mathbf{ifail}}>0$
The computation of the spectral factorization failed to converge.
${\mathbf{ifail}}=-1$
On entry, ${\mathbf{uplo}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{uplo}}=\text{'L'}$ or $\text{'U'}$.
${\mathbf{ifail}}=-2$
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}\ge 0$.
${\mathbf{ifail}}=-3$
${\mathbf{ifail}}=-4$
On entry, ${\mathbf{lda}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{lda}}\ge {\mathbf{n}}$.
${\mathbf{ifail}}=-6$
iflag was set to a nonzero value in f.
${\mathbf{ifail}}=-99$
See Section 3.9 in How to Use the NAG Library and its Documentation for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 3.8 in How to Use the NAG Library and its Documentation for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 3.7 in How to Use the NAG Library and its Documentation for further information.
## 7Accuracy
Provided that $f\left(D\right)$ can be computed accurately then the computed matrix function will be close to the exact matrix function. See Section 10.2 of Higham (2008) for details and further discussion.
## 8Parallelism and Performance
f01eff is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.
f01eff makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
The integer allocatable memory required is n, and the real allocatable memory required is approximately $\left({\mathbf{n}}+\mathit{nb}+4\right)×{\mathbf{n}}$, where nb is the block size required by f08faf (dsyev).
The cost of the algorithm is $O\left({n}^{3}\right)$ plus the cost of evaluating $f\left(D\right)$. If ${\stackrel{^}{\lambda }}_{i}$ is the $i$th computed eigenvalue of $A$, then the user-supplied subroutine f will be asked to evaluate the function $f$ at $f\left({\stackrel{^}{\lambda }}_{i}\right)$, $i=1,2,\dots ,n$.
For further information on matrix functions, see Higham (2008).
f01fff can be used to find the matrix function $f\left(A\right)$ for a complex Hermitian matrix $A$.
## 10Example
This example finds the matrix cosine, $\mathrm{cos}\left(A\right)$, of the symmetric matrix
$A= 1 2 3 4 2 1 2 3 3 2 1 2 4 3 2 1 .$
### 10.1Program Text
Program Text (f01effe.f90)
### 10.2Program Data
Program Data (f01effe.d)
### 10.3Program Results
Program Results (f01effe.r)
© The Numerical Algorithms Group Ltd, Oxford, UK. 2017 | 2,875 | 9,339 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 111, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-39 | latest | en | 0.497772 |
https://www.clutchprep.com/physics/practice-problems/94749/in-yvette-and-zack-are-driving-down-the-freeway-side-by-side-with-their-windows- | 1,618,319,335,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072366.31/warc/CC-MAIN-20210413122252-20210413152252-00410.warc.gz | 768,315,478 | 28,814 | # Problem: In , Yvette and Zack are driving down the freeway side by side with their windows down. Zack wants to toss his physics book out the window and have it land in Yvettes front seat.Ignoring air resistance, should he direct his throw outward and toward the front of the car (throw 1), straight outward (throw 2), or outward and toward the back of the car (throw 3)?
###### FREE Expert Solution
93% (240 ratings)
###### Problem Details
In , Yvette and Zack are driving down the freeway side by side with their windows down. Zack wants to toss his physics book out the window and have it land in Yvettes front seat.
Ignoring air resistance, should he direct his throw outward and toward the front of the car (throw 1), straight outward (throw 2), or outward and toward the back of the car (throw 3)? | 186 | 807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-17 | latest | en | 0.968426 |
https://brilliant.org/problems/a-planar-cut-through-an-elliptical-cylinder/ | 1,490,899,892,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218199514.53/warc/CC-MAIN-20170322212959-00546-ip-10-233-31-227.ec2.internal.warc.gz | 818,428,562 | 18,724 | # A planar cut through an Elliptical Cylinder
Geometry Level 5
$\large \dfrac{x^2}{100} + \dfrac{y^2}{225} = 1$
A right cylinder whose axis is along the $$z$$-axis, has an elliptical cross-section (in the horizontal plane) given above.
If a plane whose equation is $$\sqrt{2} x + \sqrt{2} y + 2 \sqrt{3} z = 100 \sqrt{3}$$, cuts through the cylinder, then the cutting plane and the cylinder intersect in an ellipse. Find the sum of the semi-minor and semi-major axes of this ellipse. | 145 | 487 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-13 | longest | en | 0.822533 |
http://www.docstoc.com/docs/73878476/Cost-Accounting-Marginal-Costing | 1,386,294,263,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163048970/warc/CC-MAIN-20131204131728-00013-ip-10-33-133-15.ec2.internal.warc.gz | 309,599,439 | 15,718 | # Cost Accounting Marginal Costing
Document Sample
``` TILAK MAHARASHTRA UNIVERSITY
Home Assignment – Semester IV
Marks : 80 Cost & Management Accounting Code : 71412
Note :
* The Paper consists of two sections I & II
* Attempt any 4 questions from Section I
* Questions from Section I carry equal marks
* Section II is Compulsory
Section I
Q.1 What is Costing? How does Cost Accountancy differ from Financial accounting?
(15)
OR
What is Cost accounting? What are the various techniques of Costing?
Q.2 Explain the formal of cost sheet detail. (15)
OR
What is Marginal Costing? State the significance and limitations of Marginal
costing.
Q.3. What is Budget? Explain the classifications of Budget. (15)
OR
Differentiate between Cost Control & Cost Reduction..
Q. 4. (a) Write short Notes. (Any 3) (15)
1) EOQ
2) P/U ratio & CUP Analysis
3) Flexible Budget
4) Material variations
5) Over and Under absorption of overheads.
Section II
Q. 5. Multiple choice Questions (20)
1. The cost of remuneration of stores department staff is---------------------.
a) Direct labour lost b) Indirect labour lost
2. Overhead costs are allocated or apportioned to __________
a) cost centers b) product units
3. The basis of changing overheads to production units can be _______.
a) machine hours b) labour hours
c) material or labour cost d) Any of these
4. Inventory Carrying Costs consists of ________.
a) Cost of storage of materials b) Interest on Capital blocked
c) Both d) inventory
5. Cost accounting is helpful for ____________.
a) Calculation of unit cost b) fixing price of products c) both
6. Cost unit can be _________.
a) single unit of production b) Batch of production
c) any one of these
7. Elements of costs are _____, _______ & ________.
a) Material, Labour & Expenses
b) Financial, non-financial & administrative expenses
8. Marginal cost equation is _____________________.
a) S + V = FP b) S + V = F = P
c) S – V = F + P d) S = V – F - P
9. Marginal costing is useful for _________________.
a) Cost control b) both c) Profit planning
10. Worker’s time sheet can be prepared on __________.
a) Daily basis b) weekly basis
11. Job ticket is prepared for __________.
a) Each worker b) Each job
12. P/U ration : ____________________.
Contribution Contribution
i) 100 ii) 100
Cost Sales
13. BEP is ____________________.
Fixed Cost Actual Sales- Break ---
i) ii)
Sales sales
Fixed Cost Fixed Cost
iii) iv) 100
P/U ratio P/U ratio
14. Margin of salary is ______________.
a) Actual sales – BE sales b) Actal sales – budgeted sales
15. Cash budget gives an estimate of
a) Total income and expenditure b) Total receipts & payments of cash
c) None of them
16. _________ cannot be fully ---- directly to particular job.
a) Variable cost b) Material Cost c) overhead
17. Financial accounting is done with reference to _________.
a) Wide organization b) Individual department
18. Master Budget is _____________.
a) Summary of functional budget b) Finalized profit plan
c) Summarized budget
19. Zero based budgeting is based on ___________.
a) Conventional approach b) Detrimental approach
c) Futuristic approach d) Incremental approach
20. Production budget is_____________________.
a) Forecast of total sales b) Forecast of total production units
c) Forecast of total cost.
```
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A typical experiment in a virtual space - PowerPoint PPT Presentation
A typical experiment in a virtual space. Some material is put in a container at fixed T & P .
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A typical experiment in a virtual space
• Some material is put in a container at fixed T & P.
• The material is in a thermal fluctuation, producing lots of different configurations (a set of microscopic states) for a given amount of time. It is the Mother Nature who generates all the microstates.
• An apparatus is plugged to measure an observable (a macroscopic quantity) as an average over all the microstates produced from thermal fluctuation.
P
P
T
T
microscopic states (microstates)
or microscopic configurations
P
under external constraints
(N or , V or P, T or E, etc.)
Ensemble (micro-canonical, canonical, grand canonical, etc.)
T
How would you build a model system representing
a microstate of a water boiler (L~10 cm)? N = ?
Themodynamiclimit (V →∞) and simulation
• Particles (atoms, molecules, macromolecules, spins, etc.) are confined in a finite-size cell.
• Particles are in interaction: Time taken to evaluate the interaction energy or force ~ O(N2).
• - bonded interactions (bonds, angles, torsions) to connect atoms to make molecules
• - nonbonded interactions (between distant atoms)
• Particles on the surface of the cell will experience different interactions from those in the bulk!
• The total number of particles is always « small » (with respect to NA): the fraction of
• surface particles will significantly alter the average of any observables with respect to
• the expected value in the thermodynamic limit (V →∞).
Example: simple atomic system with N particles in a simple cubiccrystal state
Ns/N ~ 6 x N2/3 / N ~ 6 / N1/3
• N = 10 x 10 x 10 = 103 : ~60% surface atoms
• - N = 104: ~30% surface atoms
• N = 105: ~13% aurfaceatoms
• N = 106: ~6% surface atoms (but bigcomputational system!)
(exact calculation: Ns = 6 x(N1/3-2)2 + 12 x (N1/3-2) + 8. For N = 103, 49% surface atoms)
Periodicboundary conditions (PBC) – Born & von Karman (1912)
(from Allen & Tildesley)
A … H: images of the cell
Celldoes not have to becubic.
• - When a particle leaves the cell, one of its images comes in.
• Images are not kept in memory: Particle position after a move is checked and « folded »
• back in the cell if necessary.
• Surface effects are removed, but the largest fluctuations are ~L (cell size).
• If the system exhibits large fluctuations (for example, near a 2nd order phase transition),
• PBC will still lead to artefacts (finite-size effects).
• - Finite-size effects can be studied by considering cells of different sizes.
Periodicboundary conditions (PBC) – Born & von Karman (1912)
of Schrödinger cat
Periodicboundary condition and nonbonded interactions
L
rc
L
usuallynon-bonded pair interaction
• 2 possibilities:
• minimum image convention: consider only nearest image of a given particle when looking
• for interacting partners. Still expensive (~N2 pairs) if the cell is large!
• - Example: cell L centered on 1, interactingwith 2 and nearest images of 3, 4 and 5
• cutoff: truncate the interaction potential at a cutoff distance rc (No interaction if the distance
• between a pair isgreaterthanrc). Sphere of radius rciscenteredeachparticle.
• - Remark: usuallyrc <= L/2 in order to satisfied the minimum image convention.
x
x
x
x
Cutoff for Long-Range Non-bonded Interactions
• Direct method (simplest)
• Interactions are calculated to a cutoff distance.
• Interactions beyond this distance are ignored.
• Leads to discontinuities in energy and derivatives.
• As a pair distance moves in and out of the cutoff
• range between calculation steps, the energy jumps.
• (since the non-bond energy for that pair is included
• in one step and excluded from the next.)
Effective potential = actual potential smoothing function S(r)
• Switching function S(r)
• = 1 for small r
• = 1 0 smoothly at intermediate r
• = 0 for large r
• Should be continuously differentiable
• (so that forces can be calculated).
• Smoothly turns off non-bond interactions
over a range of distances.
• Switching range is important.
• Upper limit = the cut-off distance.
• Too large lower limit (small spline width) Unrealistic forces may result.
• Too small lower limit The feature of the equilibrium region may be lost.
Number of non-bond interactions for a 5000-atom system
as a function of cutoff distance
vdW energy of a hexapeptide crystal as a function of cutoff distance,
which does not converge until 20 Å
Estimating Non-bonded (esp. Electrostatic) Energy for Periodic Systems: Ewald Summation
For details, read Leach (pp.324-343), Allen & Tildelsley (Ch.5), and reading materials (Kofke)
P Periodic Systems: Ewald Summationeriodicboundary condition: Implementation (2d case)
y
Ly/2
1. Real coordinates
/* (xi, yi) particle i coordinates */
if (xi > Lx/2) xi = xi – Lx;
else if (xi < -Lx/2) xi = xi + Lx;
if (yi > Ly/2) yi = yi – Ly;
else if (yi < -Ly/2) yi = yi + Ly;
i
yi
xi
-Lx/2
0
Lx/2
xi-Lx
x
-Ly/2
2. Scaled (between [-0.5,0.5]) coordinates
(better to handleanycellshape):
orthorombiccell case
#define NINT(x) ((x) < 0.0 ? (int) ((x) - 0.5) : (int) ((x) + 0.5))
sxi = xi / Lx; /* (sxi, syi) particle i scaled coordinates */
syi = yi / Ly;
sxi = NINT(sxi); /* Apply PBC */
syi = NINT(syi);
xi = sxi * Lx; /* (xi, yi) particle i folded real coordinates */
yi = syi * Ly; | 1,639 | 6,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-47 | latest | en | 0.859403 |
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5th September 2003, 02:42 PM #1 damitamit diyAudio Member Join Date: Apr 2003 Location: London, UK Adding LED to gainclone hi, finished my lm3975 gainclone and just putting the finishing touches to it. (sounds great btw, with just a few hours burn in) Want to add a blue led in the amp case (one case is psu, one case is amp). Can i just connect one leg the led to one of the dc power lines going into the chip and the other to the power ground? (with suitable resistors ofcourse) Is it gonna effect the output of the amp, if its only connected to one of the channels? thanks Amit
5th September 2003, 03:06 PM #2 pete.a diyAudio Member Join Date: Jul 2002 Location: Nottingham Adding an LED Hi, No it shouldn't make any difference to the sound of the amp, just make sure you get the polarity correct. I guess you know how to calaculate the series resistor? LED SERIES Supply voltage = 24 - operating led voltage ( normally around 3.3 v) Divided by the forward current of the led(50Ma) =24-3.3 /.05= 414 Ohms I used a 1k 2w resistor on mine, as that's what i to hand and it works fine.
5th September 2003, 03:15 PM #3 damitamit diyAudio Member Join Date: Apr 2003 Location: London, UK yep, thanks. just tryin to figure out if i have a high enough wattage resistor in the parts box. could i just use p=iv? p = (supply voltage-working voltage) * current draw p = 26-4.5 * 0.02 p= 0.43 W hmm, think i only have 1/4 W resistors about..
5th September 2003, 03:32 PM #4 lieven diyAudio Member Join Date: Aug 2002 Location: Belgium just keep some old pcb's of whatever in the nailbox, should find what you need on them.
5th September 2003, 11:20 PM #5 indoubt diyAudio Member Join Date: Aug 2003 Location: Sweet lake city And for those who are afraid to put a led on the secondary power, there is a small pcb available for a couple of € and you can feed the led from the main power. if that will affect the sound than everything op to the generator will so it will be out of your control anyway (example conrad partno:184985-44) __________________ better be indoubt untill you're sure
5th September 2003, 11:24 PM #6 Sandy H. diyAudio Member Join Date: Nov 2002 Location: Charlotte, NC - USA Another option? I recall a while back that Peter Daniel and others suggested wrapping additional windings around the outside of a torroidal transformer to create a separate power supply. If this is a viable solution, it would be practically free and unrelated to the circuit. Just trying to throw something else out there. Sandy.
6th September 2003, 02:14 AM #7 ir diyAudio Member Join Date: Aug 2003 Location: WGTN, NZ the resistor wattage is irrelevant as the LED will only be drawing typically 20mA. the formula is R=(E-Vf)x1000/I where R is the series resistance in Ohms, E is the supply voltage, Vf is the LED's forward voltage (typically 2V), I is the LED current in mA so for a normal, 5mm LED for example, running of +22V you need 1KOhm for an ultra bright blue LED, Vf=3.8V, you need 910Ohms, using 1K is fine. 25V would give 1060 use 1.1 or 1.2K 32V would give 1410 use 1.5K simply use the formula and then the nearest, HIGHER, common value i.e. 800=820, 910=1K, 1100=1200 etc (@10%tol, less tol will provide more divisions and a closer match - but again, that's totally unnecessary)
Circlotron
diyAudio Member
Join Date: Jun 2002
Location: Melbourne, Australia
Then tell me, "future boy", who is president in the United States in 1985?
Quote:
Originally posted by ir OC4Free - Performance on the cheap.
A bit off topic- sorry- but I couldn't help notice the date format on your website e.g. 20-08-2k3. We ain't there yet....
__________________
Best-ever T/S parameter spreadsheet.
http://www.diyaudio.com/forums/multi...tml#post353269
6th September 2003, 03:10 AM #9 ir diyAudio Member Join Date: Aug 2003 Location: WGTN, NZ ummmm... okay. 20-08-2k3= 20th of the 8th (august)=2003 2k=2000 and 3 makes 2003 = 2k3 just like roman numerals really MMI = 1000+1000+1 logical eh?
Nuuk
diyAudio Member
Join Date: Feb 2003
Location: Somerset, SW England
Quote:
2k=2000 and 3 makes 2003 = 2k3
Except that in 2K3, the 3 is 300! I'm not nit picking here but many less experienced may get confused when selecting resistors or capacitors specified like that.
Think of the letter like a decimal point . So 2.3 (k=thousand) = 2,300.
__________________
The truth need not be veiled, for it veils itself from the eyes of the ignorant.
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Wiki | 1,971 | 7,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-26 | latest | en | 0.909763 |
https://www.australiabesttutors.com/recent_question/65188/sit384-cyber-security-analyticspass-task-8-1p-pca-dimensionality | 1,642,632,766,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301592.29/warc/CC-MAIN-20220119215632-20220120005632-00452.warc.gz | 710,695,108 | 9,681 | ### Recent Question/Assignment
SIT384 Cyber security analytics
Pass Task 8.1P: PCA dimensionality reduction
Task description:
PCA (Principle Component Analysis) is a dimensionality reduction technique that projects the data into a lower dimensional space. It can be used to reduce high dimensional data into 2 or 3 dimensions so that we can visualize and hopefully understand the data better.
In this task, you use PCA to reduce the dimensionality of a given dataset and visualize the data.
You are given:
• Breast cancer dataset which can be retrieved from:
from sklearn.datasets import load_breast_cancer cancer = load_breast_cancer() detailed info available at: https://scikitlearn.org/stable/modules/generated/sklearn.datasets.load_breast_cancer.html
• PCA(n_components=2)
• 3D plot settings: (Please refer to prac7 for 3D plot examples) from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure(figsize=(10, 8)) cmap = plt.cm.get_cmap(-Spectral-) ax = Axes3D(fig, rect=[0, 0, .95, 1], elev=10, azim=10) ax.scatter(x,y,z, c=cancer.target, cmap=cmap)
• Other settings of your choice
You are asked to:
• use StandardScaler() to first fit and transform the cancer.data,
• apply PCA (n_components=2) to fit and transform the scaled cancer.data set
• print the scaled dataset shape and PCA transformed dataset shape for comparison
• create 2D plot with the first principal component as x axis and the second principal component as y axis
• set proper xlabel, ylabel for the 2D plot
• print the PCA component shape and component values
• create a 3D plot with the first 3 features (as x,y and z) of the scaled cancer.data set
• create a 3D plot with the first principal component as x axis and the second principal component as y axis, no value for z axis
• set proper title for the two 3D plots
Sample output as shown in the following figures are for demonstration purposes only. Yours might be different from the provided.
Submission:
Submit the following files to OnTrack:
1. Your program source code (e.g. task8_1.py)
2. A screen shot of your program running
Check the following things before submitting:
1. Add proper comments to your code
SIT384 Cyber security analytics
Pass Task 7.1P: K-Means and Hierarchical Clustering
Task description:
In machine learning, clustering is used for analyzing and grouping data which does not include prelabelled class or even a class attribute at all. K-Means clustering and hierarchical clustering are all unsupervised learning algorithms.
K- means is a collection of objects which are “similar” between them and are “dissimilar” to the objects belonging to other clusters. It is a division of objects into clusters such that each object is in exactly one cluster, not several.
In Hierarchical clustering, clusters have a tree like structure or a parent child relationship. Here, the two most similar clusters are combined together and continue to combine until all objects are in the same cluster.
In this task, you use K-Means and Agglomerative Hierarchical algorithms to cluster a synthetic dataset and compare their difference.
You are given:
• np.random.seed(0)
• make_blobs class with input:
o n_samples: 200
o centers: [3,2], [6, 4], [10, 5] o cluster_std: 0.9
• KMeans() function with setting: init = -k-means++-, n_clusters = 3, n_init = 12
• AgglomerativeClustering() function with setting: n_clusters = 3, linkage = average
• Other settings of your choice
You are asked to:
• plot your created dataset
• plot the two clustering models for your created dataset
• set the K-Mean plot with title “KMeans”
• set the Agglomerative Hierarchical plot with title “Agglomerative Hierarchical”
• calculate distance matrix for Agglomerative Clustering using the input feature matrix (linkage = complete)
• display dendrogram
Sample output as shown in the following figure is for demonstration purposes only. Yours might be different from the provided.
Submission:
Submit the following files to OnTrack:
1. Your program source code (e.g. task7_1.py)
2. A screen shot of your program running
Check the following things before submitting:
1. Add proper comments to your code | 970 | 4,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-05 | latest | en | 0.783492 |
https://math.stackexchange.com/questions/2297740/solve-cos-2x-sin-2x-sqrt-3-cos-4x | 1,653,544,625,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662601401.72/warc/CC-MAIN-20220526035036-20220526065036-00623.warc.gz | 454,154,300 | 66,191 | # Solve $\cos 2x - \sin 2x = \sqrt 3\cos 4x$
How do I solve this trig equation?
$$\cos 2x - \sin2x = \sqrt{3} \cos 4x$$
My work is,
$$\cos 2x - \sin 2x = √3(\cos 2x - \sin 2x)(\cos 2x + \sin 2x)$$ $$⇔(\cos 2x - \sin 2x)((1-√3(\cos 2x + \sin 2x))= 0$$ Then, $$\tan 2x = 1 \ \ \ \text{or} \ \ \ (\cos 2x + \sin 2x)= \dfrac1{\sqrt3}$$
I don't know whether this way is correct. Please someone help for a better solution.
• Show us the work from your different attempts. That will help us better understand any misunderstandings you might have, and what you may have overlooked. May 26, 2017 at 15:19
• I have tried writing the cos 4x as (cos 2x)^2-(sin 2x)^2 May 26, 2017 at 15:24
• It should be $\cos^2 2x - \sin^2 2x$ May 26, 2017 at 15:25
• actually, what you wrote is common short-hand and means precisely what the asker used. Nimantha: note that $(\cos 2x)^2 = \cos^2(2x)$. So you are indeed correct. May 26, 2017 at 15:27
We need to solve that $$\cos2x-\sin2x=\sqrt3(\cos2x-\sin2x)(\cos2x+\sin2x),$$ which gives $\cos2x-\sin2x=0$, which is $x=\frac{\pi}{8}+\frac{\pi}{2}k,$ $k\in\mathbb Z$ or $$\cos2x+\sin2x=\frac{1}{\sqrt3},$$ which is $$\cos\left(2x-\frac{\pi}{4}\right)=\frac{1}{\sqrt6},$$ which gives $x=\frac{\pi}{8}\pm\frac{1}{2}\arccos\frac{1}{\sqrt6}+\pi k$.
Use $\cos2y=\cos^2y-\sin^2y$ in $\cos4x=\cos2(2x)$
Also, for $\sqrt3(\cos2x+\sin2x)=1,$ $$\cos\left(2x-\dfrac\pi4\right)=\dfrac1{\sqrt6}$$
$$\implies2x-\dfrac\pi4=2m\pi\pm\arccos\dfrac1{\sqrt6}$$ where $m$ is any integer. | 648 | 1,498 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2022-21 | longest | en | 0.757602 |
http://www.ssc.education.ed.ac.uk/BSL/maths/bslcubex.html | 1,548,019,369,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583739170.35/warc/CC-MAIN-20190120204649-20190120230649-00182.warc.gz | 392,990,239 | 2,782 | ## BSL Maths Glossary - cube root - Example (symbol ³√)
Subject level: Credit
Example
This symbol is called a cube root. Let's put in a number. So the cube root of 8 equals what? How do we work this out? I'll show you. You need to have exactly the same number repeated, not different ones. They must be the same. So 2x2 is 4 and that times 2 is 8. So that is three 2s. So the cube root of 8 is 2. So here you would put 2.
Here's one more. The cube root of 1000. That's a difficult one. How can we work this out? Remember we need to have the same number repeated. You have to work out that we need to have three numbers.
5 x 5 is 25 x 5 - not that wouldn't work.
We need to find 1000. What about 10 x 10 x10 - oh that might work. So 10x10 is 100 and x 10 is 1000.x
So that means you have used the same number three times and that's right. So here you would put 10 (as the answer). | 261 | 886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-04 | latest | en | 0.975891 |
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jvmoore1
Newbie
Joined: 10 Feb 2006
Posts: 4
Posted: Fri Feb 10, 2006 7:05 am Post subject: Inventory Control I have searched the forums but so far nothing seems to help..of course i am a n00b so this should be expected... We receive 5 trucks a day with 30 rolls per truck. Each roll has its own number (i.e 3160-3190). so to input all i have is date roll number receiver (example: 1. 1-30-06 31360 m-2162 2. 1-30-06 31361 m-2162) there will be 15000 rolls so i dont really want to do the to pull these rolls out. so my thought (which can be corrected) was to make sheet 2 the outbound sheet and somehow find a rule so whenever i type the roll number in sheet 2 it will be pulled from sheet 1. can this be done? remember i am a n00b so treating me like a k-5 child is acceptable!
Villeroy
Super User
Joined: 04 Oct 2004
Posts: 10106
Location: Germany
Posted: Fri Feb 10, 2006 8:04 am Post subject: The most simple solution requires the Roll-column to be the first one Put the roll-number in Sheet2.A1 Sheet2.A2: =VLOOKUP(\$A\$1;\$Sheet1.\$A\$1:\$C\$65536;2;0) Looks up roll-number in first col of A1:C65536 and gives the value of the second column. Sheet2.A3: =VLOOKUP(\$A\$1;\$Sheet1.\$A\$1:\$C\$65536;3;0) Looks up roll-number in first col of A1:C65536 and gives the value of the third column. If A1 has a roll-number, not existing in first col of sheet1, you get #NA as result. If (by mistake) there is more than one item with the specified number, you get the first item found. So you may use: sheet2.B1 =COUNTIF(sheet1.\$A\$1:\$A\$65536;\$A\$1) just to indicate duplicate items.
jvmoore1
Newbie
Joined: 10 Feb 2006
Posts: 4
Posted: Fri Feb 10, 2006 9:03 am Post subject:
Villeroy wrote: The most simple solution requires the Roll-column to be the first one Put the roll-number in Sheet2.A1 Sheet2.A2: =VLOOKUP(\$A\$1;\$Sheet1.\$A\$1:\$C\$65536;2;0) Looks up roll-number in first col of A1:C65536 and gives the value of the second column. Sheet2.A3: =VLOOKUP(\$A\$1;\$Sheet1.\$A\$1:\$C\$65536;3;0) Looks up roll-number in first col of A1:C65536 and gives the value of the third column. If A1 has a roll-number, not existing in first col of sheet1, you get #NA as result. If (by mistake) there is more than one item with the specified number, you get the first item found. So you may use: sheet2.B1 =COUNTIF(sheet1.\$A\$1:\$A\$65536;\$A\$1) just to indicate duplicate items.
ok
so i changed the layout to be roll number receiver date
A1 B1 C1
31360 m-2162 1-30-06
so on sheet 2 i posted the rule and it seems to only work for whatever roll i put in A1...
i tried dragging the rule in B2 down, but intstead of it changing the cell for teh rule it make the exact same rule (makes sense?)
also, when i type the roll number in A1 of sheet 2 the roll number in sheet 1 is still there, i need to remove this once it is in sheet 2
thank you so much for what you have done so far...
Villeroy
Super User
Joined: 04 Oct 2004
Posts: 10106
Location: Germany
Posted: Fri Feb 10, 2006 9:46 am Post subject:
Quote: so my thought (which can be corrected) was to make sheet 2 the outbound sheet and somehow find a rule so whenever i type the roll number in sheet 2 it will be pulled from sheet 1.
We get what you requested so far.
Put the first roll-number in Sheet2.A1
Sheet2.B1: =VLOOKUP(\$A1;\$Sheet1.\$A\$1:\$C\$65536;2;0)
Looks up roll-number in first col of A1:C65536 and gives the value of the second column.
Sheet2.C1: =VLOOKUP(\$A1;\$Sheet1.\$A\$1:\$C\$65536;3;0)
Looks up roll-number in first col of A1:C65536 and gives the value of the third column.
sheet2.D1 =COUNTIF(sheet1.\$A\$1:\$A\$65536;\$A1)
I put all formulas referring to the lookup-number into one row and the slight -but important - difference: I refer to cell \$A1 instead of \$A\$1.
Read the \$ as "exactly ..."
From the view point of cell B1, \$A1 is the value in the same row, but "exactly in" column A.
Copy down B1:D1 and you'll see how this rule is applied.
As a test, copy B1:D1 to the right and you'll see why I prefer keeping the \$ in front of the A. It makes things easier to reorganize.
Now you may type a roll-number in col A and get the lookup-values and count in B to D.
A spreadsheet handles your data "as is". There is no automatic which can remove anything automaticly (some advanced builtin mechanisms are able to add data). This could be implemented with some addon-program, often called a "macro".
It seems, you have a list of things to be done on sheet1, but I'm not shure about what kind of information you want to get on sheet2.
Anyway, I have another formula for column E:
This produces a hyperlink to the cell where the value of A1 is found in col A of sheet1. Notice that quoted string "#sheet1." as first part of the hyperlink address is "frozen". Unlike the other references to sheet1 it won't be updated in case you rename sheet1. You will have to use search/replace then.
Villeroy
Super User
Joined: 04 Oct 2004
Posts: 10106
Location: Germany
Posted: Fri Feb 10, 2006 10:09 am Post subject: After rereading It's obvious that you want a list about what is done. Formula in some column of sheet1: =ISNUMBER(MATCH(\$A1;\$sheet2.\$A\$1:\$A\$65536;0)) gives TRUE if there is an entry in sheet2. Or =IF(ISNUMBER(MATCH(\$A1;\$sheet2.\$A\$1:\$A\$65536;0));"Done";"To Do") Now you may select entire used columns of sheet1 and call Menu:Data>Filter>Auto Filter
jvmoore1
Newbie
Joined: 10 Feb 2006
Posts: 4
Posted: Fri Feb 10, 2006 12:18 pm Post subject:
Villeroy wrote: After rereading It's obvious that you want a list about what is done. Formula in some column of sheet1: =ISNUMBER(MATCH(\$A1;\$sheet2.\$A\$1:\$A\$65536;0)) gives TRUE if there is an entry in sheet2. Or =IF(ISNUMBER(MATCH(\$A1;\$sheet2.\$A\$1:\$A\$65536;0));"Done";"To Do") Now you may select entire used columns of sheet1 and call Menu:Data>Filter>Auto Filter
awesome!! everything works great excpet for the "to do" "done" part...each cell says "to do" even though i have that cell on sheet 2...
but even if that doesnt work this is great...i appreciate it soo much
Villeroy
Super User
Joined: 04 Oct 2004
Posts: 10106
Location: Germany
Posted: Fri Feb 10, 2006 12:39 pm Post subject: Mmmmh, Do the cell of the "done"-function refer to the according lookup-value? Hit F2 (edit-mode) and you get some colored borders, indicating the referred cells on the same sheet. The first formula should refer to the first roll-number.
jvmoore1
Newbie
Joined: 10 Feb 2006
Posts: 4
Posted: Fri Feb 10, 2006 1:11 pm Post subject:
Villeroy wrote: Mmmmh, Do the cell of the "done"-function refer to the according lookup-value? Hit F2 (edit-mode) and you get some colored borders, indicating the referred cells on the same sheet. The first formula should refer to the first roll-number.
yeap that was it...
thank you so much!!! i appreciate it....seriously...
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# How many seconds are in summer?
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The duration of Summer Evening is 1800.0 seconds. | 410 | 1,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.943372 |
https://codereview.stackexchange.com/questions/180273/find-palindromic-substrings-as-efficiently-as-possible/180323 | 1,566,235,787,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314852.37/warc/CC-MAIN-20190819160107-20190819182107-00522.warc.gz | 424,095,812 | 33,345 | # Find palindromic substrings as efficiently as possible
I've written a function to calculate all the distinct substrings of a given string, that are palindromes. The string is passed as the argument to the function. For example, the string abba is a palindrome because reversed it is also abba. If the function takes the string aabaa it returns 5, because the distinct substrings are a, aa, aabaa, aba, b.
I've currently made use of two for loops for this to work, however, when the string gets big, it's very inefficient due to the fact it iterates one by one. Any suggestions as to how I can make this significantly more efficient? Recursion maybe?
The function:
function countPalindromesInString(s) {
let subStrings = [];
for (let i = 0; i < s.length; i++) {
for(let j = 0; j < s.length - i; j++) {
let subString = s.substring(j, j + i + 1);
if(subString === subString.split('').reverse().join('') && !subStrings.includes(subString)) {
subStrings.push(subString);
}
}
}
return subStrings.length;
}
• Is the full string guaranteed to be a palindrome? – Gerrit0 Nov 13 '17 at 4:59
• @Gerrit0 yes the initial input is a palindrome. – Matt Kent Nov 13 '17 at 9:26
• This can be solved in O(n) either with Manchester or palindromic tree : adilet.org/blog/25-09-14 – juvian Nov 13 '17 at 16:16
• @juvian Better to post a short answer than review the code in comments, which are temporary. – dcorking Nov 17 '17 at 8:50
• I have posted your review as community wiki (so I don't get credit for your work): it is definitely a brief review, not merely a link. If you post in your own name, we can delete my answer. – dcorking Nov 18 '17 at 9:01
You code looks pretty neat, i do not think that using recursion would improve it very much, but i have some minor suggestions;
### 1) Performance-wise improvements
The only thing that seems odd at first glance is how you reverse the string to compare subString.split('').reverse().join('');
I had to look it up, but i found that the way that you are doing its the right way using In-built functions, and its pretty much good for almost all cases, but if you want to improve a little the performance you could use something like:
function reverse(s) {
var o = [];
for (var i = 0, len = s.length; i <= len; i++)
o.push(s.charAt(len - i));
return o.join('');
}
Using this function is has an improve in performance over your current implementation.
You could extract your conditional and use it as a function so your intent would be more clear:
let isPalindrome = function (word, words) {
return word === word.split('').reverse().join('') && !words.includes(word)
}
and use this function in the conditional as:
if(isPalisdrome(subString, subStrings)) {
subStrings.push(subString);
}
### 3) Conclusion
In the end putting all together the code would be something like:
function reverse(s) {
var o = [];
for (var i = 0, len = s.length; i <= len; i++)
o.push(s.charAt(len - i));
return o.join('');
}
function isPalindrome(word, words) {
return word === reverse(word) && !words.includes(word)
}
function countPalindromesInString(s) {
let subStrings = [];
for (let i = 0; i < s.length; i++) {
for(let j = 0; j < s.length - i; j++) {
let subString = s.substring(j, j + i + 1);
if(isPalindrome(subString, subStrings)) {
subStrings.push(subString);
}
}
}
return subStrings.length;
}
Checking for a duplicate in the results array is one slow point. It adds one level of complexity because each contains call has to loop over the whole list. Instead use a Set (if you can use ECMAScript 6) or store the strings as a key in an object (this however requires counting the keys at the end which can be a bit bothersome unless you can use Object.keys() which in turn requires ECMAScript 5.1):
var resultSet = {};
// ... inside the loop if the substring is a palindrome
resultSet[substring] = true;
var count = 0;
for (key in resultSet) {
if (resultSet.hasOwnProperty(key)) count++;
}
return count;
(I just realized you are using let so you can use Set, too. I'm just leaving this in, in case it interests someone else.)
There is no reason to actually create the reverse string in order to check for a palindrome. Instead simply loop over the first half of the string and compare the nth character with the (length - n - 1)th character.
function isPalindrom(str) {
var len = str.length;
var half = Math.floor(len / 2);
for (var i = 0; i < half; i++) {
if (str.charAt(i) !== str.charAt(len - i - 1)) {
return false;
}
}
return true;
}
This even could be taken one more step: There is no reason to create the substring for the palindrome detection, except to put it in the result set afterwards.
function isPalindrom(str, from, to) {
let len = from - to + 1;
let half = Math.floor(len / 2);
let end = from + half + 1;
for (var i = from; i < end; i++) {
if (str.charAt(i) !== str.charAt(len - i - 1)) {
return false;
}
}
return true;
}
function countPalindromesInString(s) {
let subStrings = new Set();
for (let i = 0; i < s.length; i++) {
for (let j = 0; j < s.length - i; j++) {
let end = j + i + 1;
if (isPalindrom(s, j, end)) {
subStrings.push(s.substring(j, end));
}
}
}
return subStrings.size;
}
You should also take into consideration the JS methods when calculating time and space complexity. It's not as simple as $$\O(n^2)\$$ time because you used two nested for loops. Slice, Split, reverse, join are all also $$\O(n)\$$ operations under the hood. Split also creates a brand new array so that is also $$\O(n)\$$ space.
Originally posted by user Juvian:
This can be solved in O(n) either with Manchester or palindromic tree : adilet.org/blog/25-09-14
Both manchester and palindromic tree are a bit harder to implement and definitely not easy to understand, so as long as performance is not needed its good to avoid them. | 1,508 | 5,791 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-35 | latest | en | 0.875049 |
http://www.britannica.com/EBchecked/topic/127341/combinatorics/21876/Combinatorics-during-the-20th-century | 1,430,866,181,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430457008123.23/warc/CC-MAIN-20150501051008-00027-ip-10-235-10-82.ec2.internal.warc.gz | 290,897,758 | 17,372 | # Combinatorics
Mathematics
Alternate title: combinatorial mathematics
## Combinatorics during the 20th century
Many factors have contributed to the quickening pace of development of combinatorial theory since 1920. One of these was the development of the statistical theory of the design of experiments by the English statisticians Ronald Fisher and Frank Yates, which has given rise to many problems of combinatorial interest; the methods initially developed to solve them have found applications in such fields as coding theory. Information theory, which arose around midcentury, has also become a rich source of combinatorial problems of a quite new type.
Another source of the revival of interest in combinatorics is graph theory, the importance of which lies in the fact that graphs can serve as abstract models for many different kinds of schemes of relations among sets of objects. Its applications extend to operations research, chemistry, statistical mechanics, theoretical physics, and socioeconomic problems. The theory of transportation networks can be regarded as a chapter of the theory of directed graphs. One of the most challenging theoretical problems, the four-colour problem (see below) belongs to the domain of graph theory. It has also applications to such other branches of mathematics as group theory.
The development of computer technology in the second half of the 20th century is a main cause of the interest in finite mathematics in general and combinatorial theory in particular. Combinatorial problems arise not only in numerical analysis but also in the design of computer systems and in the application of computers to such problems as those of information storage and retrieval.
Statistical mechanics is one of the oldest and most productive sources of combinatorial problems. Much important combinatorial work has been done by applied mathematicians and physicists since the mid-20th century—for example, the work on Ising models (see below The Ising problem).
In pure mathematics, combinatorial methods have been used with advantage in such diverse fields as probability, algebra (finite groups and fields, matrix and lattice theory), number theory (difference sets), set theory (Sperner’s theorem), and mathematical logic (Ramsey’s theorem).
In contrast to the wide range of combinatorial problems and the multiplicity of methods that have been devised to deal with them stands the lack of a central unifying theory. Unifying principles and cross connections, however, have begun to appear in various areas of combinatorial theory. The search for an underlying pattern that may indicate in some way how the diverse parts of combinatorics are interwoven is a challenge that faces mathematicians in the last quarter of the 20th century.
## Binomial coefficients
An ordered set a1, a2,…, ar of r distinct objects selected from a set of n objects is called a permutation of n things taken r at a time. The number of permutations is given by nPn = n(n − 1)(n − 2)⋯ (nr + 1). When r = n, the number nPr = n(n − 1)(n − 2)⋯ is simply the number of ways of arranging n distinct things in a row. This expression is called factorial n and is denoted by n!. It follows that nPr = n!/(nr)!. By convention 0! = 1.
A set of r objects selected from a set of n objects without regard to order is called a combination of n things taken r at a time. Because each combination gives rise to r! permutations, the number of combinations, which is written (n/r), can be expressed in terms of factorials
.
The number (n/r) is called a binomial coefficient because it occurs as the coefficient of prqnr in the binomial expansion—that is, the re-expression of (q + p)n in a linear combination of products of p and q
.
in the binomial expansion is the probability that an event the chance of occurrence of which is p occurs exactly r times in n independent trials (see probability theory).
The answer to many different kinds of enumeration problems can be expressed in terms of binomial coefficients. The number of distinct solutions of the equation x1 + x2 +⋯+ xn = m, for example, in which m is a non-negative integer mn and in which only non-negative integral values of xi are allowed is expressible this way, as was found by the 17th–18th-century French-born British mathematician Abraham De Moivre
.
## Multinomial coefficients
If S is a set of n objects and if n1, n2,…, nk are non-negative integers satisfying n1 + n2 +⋯+ nk = n, then the number of ways in which the objects can be distributed into k boxes, X1, X2,…, Xk, such that the box Xi contains exactly ni objects is given in terms of a ratio constructed of factorials
.
This number, called a multinomial coefficient, is the coefficient in the multinomial expansion of the nth power of the sum of the {pi}
If all of the {pi} are non-negative and sum to 1 and if there are k possible outcomes in a trial in which the chance of the ith outcome is pi, then the ith summand in the multinomial expansion is the probability that in n independent trials the ith outcome will occur exactly ni times, for each i, 1 i k.
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Topic: Matheology § 223: AC and AMS
Replies: 102 Last Post: Apr 18, 2013 12:26 AM
Messages: [ Previous | Next ]
mueckenh@rz.fh-augsburg.de Posts: 18,076 Registered: 1/29/05
Re: Matheology § 223: AC and AMS
Posted: Mar 16, 2013 2:00 PM
On 16 Mrz., 18:17, fom <fomJ...@nyms.net> wrote:
>
> > 2) Do you agree that choosing a number from a set with more than 1
> > element means writing or speaking or at least thinking the name of the
> > number?
>
> No. The use of logic and axioms is justifiable as
> representations that formalize mathematical practice.
The practice must not become unpracticable by logic.
> They are normative ideals against which mathematical
> practice is measured.
Logic and formalization *describe* practice, they cannot change it.
> Your question applies to the faithfulness of those
> representations. What is "nameable in principle" may
> not be materially nameable.
Here is the question whether something can be chosen, not whether it
can be "in priciple" chosen.
>
> > 5) Zermelo's AC requires that uncountably many names can be written,
> > said or thought.
>
> No. Zermelo's AC requires that one name can be written
> with certainty.
>
> "the cartesian product of non-empty sets is non-empty"
You could with same ease write: Fermat's last theorem can be violated
with certainty. What would be the difference?
>
> > 6) In this respect it resembles the statement that a second prime
> > number triple beyond (3, 5, 7) can be found, perhaps even infinitely
> > many.
>
> My unfamiliarity with number theory,
Of six successive naturals, at least two are divisble by 3, one of
them necessarily being an odd one. Therefore there cannot be another
prime triple. But since you refrain from arguing and adhere to
provably false claims, if given the form of axioms, you could also
accept this one.
Regards, WM
Date Subject Author
3/14/13 Alan Smaill
3/14/13 mueckenh@rz.fh-augsburg.de
3/14/13 Virgil
3/14/13 fom
3/14/13 mueckenh@rz.fh-augsburg.de
3/14/13 fom
3/14/13 mueckenh@rz.fh-augsburg.de
3/14/13 fom
3/15/13 mueckenh@rz.fh-augsburg.de
3/15/13 fom
3/15/13 mueckenh@rz.fh-augsburg.de
3/15/13 Virgil
3/15/13 fom
3/16/13 mueckenh@rz.fh-augsburg.de
3/16/13 fom
3/16/13 mueckenh@rz.fh-augsburg.de
3/16/13 fom
3/16/13 mueckenh@rz.fh-augsburg.de
3/16/13 Virgil
3/17/13 fom
3/17/13 Virgil
3/16/13 mueckenh@rz.fh-augsburg.de
3/16/13 Virgil
3/17/13 fom
3/17/13 mueckenh@rz.fh-augsburg.de
3/17/13 Virgil
3/17/13 mueckenh@rz.fh-augsburg.de
3/17/13 Virgil
3/18/13 fom
3/18/13 mueckenh@rz.fh-augsburg.de
3/18/13 fom
3/18/13 mueckenh@rz.fh-augsburg.de
3/18/13 fom
3/18/13 Virgil
3/18/13 mueckenh@rz.fh-augsburg.de
3/18/13 fom
3/18/13 mueckenh@rz.fh-augsburg.de
3/18/13 fom
3/19/13 mueckenh@rz.fh-augsburg.de
3/19/13 fom
3/19/13 fom
3/19/13 mueckenh@rz.fh-augsburg.de
3/19/13 fom
3/19/13 Virgil
3/19/13 fom
3/19/13 Virgil
3/19/13 Virgil
4/17/13 Virgil
3/18/13 Virgil
3/18/13 Virgil
3/18/13 fom
3/18/13 fom
3/18/13 mueckenh@rz.fh-augsburg.de
3/18/13 fom
3/18/13 Virgil
3/19/13 fom
3/18/13 Virgil
3/18/13 fom
3/18/13 fom
3/18/13 mueckenh@rz.fh-augsburg.de
3/18/13 fom
3/18/13 mueckenh@rz.fh-augsburg.de
3/18/13 Virgil
3/18/13 fom
3/18/13 fom
3/18/13 mueckenh@rz.fh-augsburg.de
3/18/13 Virgil
3/19/13 mueckenh@rz.fh-augsburg.de
3/19/13 Virgil
3/19/13 mueckenh@rz.fh-augsburg.de
3/19/13 Virgil
3/18/13 fom
3/19/13 mueckenh@rz.fh-augsburg.de
3/19/13 Virgil
3/19/13 fom
4/17/13 Virgil
4/18/13 fom
3/18/13 Virgil
3/18/13 mueckenh@rz.fh-augsburg.de
3/18/13 Virgil
3/18/13 Virgil
3/18/13 Virgil
3/16/13 Virgil
3/16/13 Virgil
3/17/13 fom
3/15/13 fom
3/16/13 mueckenh@rz.fh-augsburg.de
3/16/13 Virgil
3/15/13 Virgil
3/15/13 mueckenh@rz.fh-augsburg.de
3/15/13 Virgil
3/15/13 fom
3/15/13 fom
3/15/13 Virgil
3/15/13 fom
3/16/13 Virgil
3/14/13 Virgil
3/14/13 Virgil
3/16/13 mueckenh@rz.fh-augsburg.de
3/16/13 Virgil
3/17/13 fom | 1,610 | 4,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-22 | longest | en | 0.924976 |
https://1st-in-babies.com/600-grams-is-how-many-ounces-update-new/ | 1,679,857,797,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946445.46/warc/CC-MAIN-20230326173112-20230326203112-00432.warc.gz | 101,780,335 | 49,334 | 600 Grams Is How Many Ounces? Update New
# 600 Grams Is How Many Ounces? Update New
Let’s discuss the question: 600 grams is how many ounces. We summarize all relevant answers in section Q&A of website 1st-in-babies.com in category: Blog MMO. See more related questions in the comments below.
## How much is 600 grams in a cup?
Basic ingredients
Product Density Grams in 1 cup (US)
Water 1000 236.59
Flour 600 141.60
Milk 1030 243.08
Sugar 845 199.42
Apr 6, 2022
## How many pounds ounces is 600 grams?
Converting between grams and pounds
Grams Pounds Pounds and ounces
600 grams 1.323 lb 1 lb, 5.16 oz
700 grams 1.543 lb 1 lb, 8.69 oz
800 grams 1.764 lb 1 lb, 12.22 oz
900 grams 1.984 lb 1 lb, 15.75 oz
### ✅ How Many Grams In An Ounce
✅ How Many Grams In An Ounce
✅ How Many Grams In An Ounce
## How many grams is a full Oz?
There are 28 grams within one ounce. If you can remember this number, even if you find yourself without this handy cooking conversion chart, you’ll be able to make some quick calculations.
## Is 250g 8 oz?
Weight
Grams Pounds/ounces
225g 8oz
250g 9oz
300g 10oz
350g 12oz
## How much grams are in a cup?
Dry Goods
Cups Grams Ounces
1/2 cup 64 g 2.25 oz
2/3 cup 85 g 3 oz
3/4 cup 96 g 3.38 oz
1 cup 128 g 4.5 oz
Nov 19, 2020
## How do I convert grams to cups?
To convert a gram measurement to a cup measurement, divide the weight by 236.588236 times the density of the ingredient or material. Thus, the weight in cups is equal to the grams divided by 236.588236 times the density of the ingredient or material.
## What item is 600grams?
A pair of blue jeans
A pair of blue jeans is that clothing item you can wear with almost anything, ranging from suits, shirts, tee shirts, jackets, and a whole lot more. A pair of jeans for an adult male weighs around 600 grams and it becomes heavier when soaked in water.
## What is a gram equal to in weight?
1 gram (g) is equal to 0.00220462262185 pounds (lbs).
## Are pounds and ounces the same?
1 pound (lb) is equal to 16 Ounces (oz).
## How many grams is 8fl oz?
Ounces to Grams conversion table
Ounces (oz) Grams (g) Kilograms+Grams (kg+g)
8 oz 226.80 g 0 kg 226.80 g
9 oz 255.15 g 0 kg 255.15 g
10 oz 283.50 g 0 kg 283.50 g
20 oz 566.99 g 0 kg 566.99 g
## Which is more 1 oz or 1g?
If you’re wondering how an ounce compares to a gram, it turns out that 1 ounce is a lot more mass than 1 gram. In fact, 1 ounce is approximately equal to 28.35 grams.
## What is 14 fluid ounces in grams?
### 1 oz how many gram
1 oz how many gram
1 oz how many gram
## What is two cups in grams?
Brown sugar
AMOUNT IN US CUPS AMOUNT IN GRAMS AMOUNT IN OUNCES
3/4 cup 150 g 5.3 oz
7/8 cup 175 g 6.2 oz
1 cup 200 g 7.1 oz
2 cups 400 g 14.1 oz
## What does 1oz mean?
ounce 1. / (aʊns) / noun. a unit of weight equal to one sixteenth of a pound (avoirdupois); 1 ounce is equal to 437.5 grains or 28.349 gramsAbbreviation: oz.
## Is 200g 8 oz?
Basic ounces to grams weight conversions
1/2 oz 15g
5 oz 140g
6 oz 170g
7 oz 200g
8 oz 225g
## How many cups is 500 grams?
Water
WATER – GRAMS TO CUPS
Grams Cups
300g 1¼ cups
400g 1½ cups + 3 tbsp
500g 2 cups + 1 tbsp
Sep 20, 2018
## How many grams is two tablespoons?
Dry Measure Equivalents
2 tablespoons 1/8 cup 28.3 grams 4 tablespoons 1/4 cup 56.7 grams 5 1/3 tablespoons 1/3 cup 75.6 grams 8 tablespoons 1/2 cup 113.4 grams 12 tablespoons 3/4 cup .375 pound
## How many grams is .5 cups?
Sugar (granulated)
Cups Grams Ounces
1/4 cup 50 g 1.77 oz
1/3 cup 67 g 2.36 oz
1/2 cup 101 g 3.55 oz
2/3 cup 134 g 4.73 oz
## How many grams is 3 cups flour?
Three cups of all purpose flour weighs 384 grams.
## How much is a gram in baking?
Baking Conversion Table
U.S. Metric
.035 ounce 1 gram
0.5 oz. 14 grams
1 oz. 28 grams
1/4 pound (lb) 113 grams
## How much in grams is 1 cup of butter?
Butter measurement equivalents
US Cups Grams Ounces
¾ cup of Butter 170.1 g 6 oz
7/8 cup of Butter 198.5 g 7 oz
1 cup of Butter 226.8 g 8 oz
## What things weigh 700 grams?
For you to stop wondering, here is a list of things that weigh 700 grams.
11 Things That Weight Around 750 Grams (g)
• Sliced bread. One slice of white bread weighs 25 grams. …
• Butter. …
• Oranges. …
• Chocolate bars. …
• 750 ml of water. …
• 2 Cans of Coke. …
• Kittens. …
• Bullfrogs.
Oct 8, 2021
### [EASY] How to Convert OUNCES to GRAMS. Ounce to Gram Conversion (oz-g)
[EASY] How to Convert OUNCES to GRAMS. Ounce to Gram Conversion (oz-g)
[EASY] How to Convert OUNCES to GRAMS. Ounce to Gram Conversion (oz-g)
## What household item weighs 400 grams?
2 Cups of granulated sugar
One of the most common baking items is granulated sugar, which weighs around 200 grams a cup. Two cups will, therefore, weigh 400 grams.
## What weighs about 1 kg?
Paper clips are an easy item to use when comparing weights. A small box of 100 paper clips weighs around 100 grams therefore 10 boxes would weigh 1000 grams which is equal to 1 kilogram. 1 regular-sized paper clip weighs 1 gram so you would need 1000 of them to equal 1 kilogram.
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## Information related to the topic 600 grams is how many ounces
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(1) An instructor obsessed with the metric system insists that all multiple-choice questions have 10 different possible answers and only one is correct. What is the probability of answering correctly if a random answer is picked?
An event is unusual if probability is 0.05 or less, so is it unusual to answer the question by correctly guessing?
(2) There are 38 slots on a roulette wheel. One slot is 0 and another slot is 00, with the other slots numbered 1-36. If you bet all your money on the number 13 for one spin, what is the probability that you will win?
Is it unusual to win when you bet on a single number when unusual has a probability of 0.05 or less?
(3) 400 randomly selected flights showed 344 arrived on time. What is the estimated probability on one of the 400 flights arriving late?
It is unusual for one of these flights to arrive late if the probability is 0.05 or less?
#### Solution Preview
An instructor obsessed with the metric system insists that all multiple-choice questions have 10 different possible answers and only one is correct. What is the probability of answering correctly if a random answer is picked.
Event is unusual if probability is 0.05 or less, so is it unusual to answer the question by correctly guessing?
If you ...
#### Solution Summary
Three questions about probability is examined.
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https://placement.freshersworld.com/infosys-placement-papers/whole-testpaper/33136251 | 1,657,187,480,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104690785.95/warc/CC-MAIN-20220707093848-20220707123848-00540.warc.gz | 488,785,620 | 53,313 | | 13396
# Whole-Testpaper
INFOSYS PAPER ON 13th JULY 2006 AT JAMIA MILLIA ISLAMIA
Hai,
Im Isha, i took the infosys test and thankfully i got through. I owe it to these sites and people who take the pain of recalling the papers n submitting them for the benefit of others. I would definitely want to pay back
1. aptitude test - puzzles (must see the papers on net n also books by shakuntala devi "puzzles 2 puzzle u" and "some more puzzles")
2. HR round (normal hr questions , we were asked general knowledge questions also, and ofcourse puzzles)
Here's the paper
There were 10 questions , some were 3 marks, some 5 marks n some 8 marks
time allowed- 1 hour
Q:1. if 40 cows can graze a field in 40 days , leaving some portion ungrazed. if 30 cows can graze the entire field in 60 days . then in how many days can 20 cows graze the same field?
Ans 1. 90 days (probably)
explanantion - 30 cows graze it in 60 days
so, 1 cow can graze in 1800 days
so, 20 will do in 90 days
Q:2. complete the series
1,2,3,8,__,224
ans . 27
1*1+1=2
2*1+1=3
3*2+2=8
8*3+3=27
27*8+8=224
1,2,3,5,16,__
ans. 79
1*2+1=3
2*3-1=5
3*5+1=16
16*5-1=79
Q:3. in how many ways can the numbers be arranged on a dice such that 1 and 6 are on opposite faces, similarly 2 and 5 , and 3 and 4 are on opp faces
Ans 3. i am not too sure , but according 2 me the answer is
6*1*4*1*2*1=48 ways
coz the first number can be placed in 6 ways
then second in just 1 way (coz it has 2 be on the opposite side)
3rd number can be placed in 4 ways
and so on........
Q:4 my mother gave me money to buy stamps of price 2paisa, 7 paisa,15 paisa, 10paisa and 20 paisa. I had to buy 5 each of three types and 6 each of the other 2 types . But on my way to the post office i forgot how many of stamps of each type were to be brought . My mother had given me rupees 3 . So i had no problem in finding out the exact amount of each one . Can you tell me which stamps were 5 in number , n whic were 6 in number
Ans . 5 stamps each of 2paisa, 7 paisa, 15 paisa
6 stanps each of 10 paisa , 20 paisa
Q:5 A man travelled a certain distance at the rate of 15 miles an hour and came back at the rate of 10 miles an hour. What is his average speed ?
Ans . 12 miles an hour
Q:6. XYZ and XYZ
+AB -AB
__________ _______________
C D E F BGA
then find the value of X,Y,Z ,G
Ans6.
945 945
+78 -78
____________ _____________
1023 867
X - 9
Y - 4
Z - 5
G - 6
Q:7. J and K are playing a match. if J wins the match , then till now he would have won twice the number of matches won by K . If K wins , then he would win the same no. of matches as K. Find the number of matches they have each won till now
Ans.7 till now J has won 3 matches and K has won 2 matches
Q:8. A very long question about " A is younger than B , but elder than C" ...........so on. It was pretty easy, You can easily figure out the answer.
Q:9. I bought 2 nursery drapes for less than 10 \$ . If i bought as many yards of each as there were pieces per yard in cents . I paid 215 cents more for one of them . Then how many yards of each did i buy ?
Q:10.
Salad problem.. Four girls Robin,Mandy,Stacy,Erica of four families Miller,Jacob,Flure,and Clark prepare four salads using the fruits Apples cherries bananas,grapes.Each girls uses 3 fruits in her salad.No body have the same combination.
1 )Robin not a Miller girl uses apples.
2) Miller and Mandy uses apples and cherries.
3) Clark uses cherries and grapes but Flure uses only one of them.
4)Erica is not Clark nor Flure.htt
1. Which is robins family:
a. Miller b. Jacob c.Flure d.Clark
2. Which fruit is not used by Mandy? aa
a. Cherries
b. Grapes
c. Apples
d.Bananas
3. Which is the combination by Erica?
a. Apples, cherries, Bananas
b. Apples ,Cherries, Grapes
c. Apples,Grapes,Bananas
d. Cherries, Grapes, Bananas
4. Which is robin's fruit combination?
a. Apples, cherries, Bananas
b. Apples ,Cherries, Grapes
c. Apples,Grapes,Bananas
d. Cherries, Grapes, Bananas
feedback | 1,327 | 4,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2022-27 | latest | en | 0.907549 |
https://brainybiker.com/combining-24-inch-wheels-and-a-20-inch-frame-is-just-wrong/ | 1,725,855,430,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00576.warc.gz | 131,526,489 | 19,427 | ## Combining 24-inch Wheels and a 20-inch Bicycle Is Just Wrong
Mini Info Bomb: In most cases, 24-inch wheels will be too big for a 20-inch fork and frame, especially if the wheel is equipped with a large tire. Thus, it’s best to avoid this conversion as it will more than likely not work.
In a rare scenario, the wheel may fit. However, if the bike uses rim brakes, they will no longer be operational.
Also, the new wheel will change the geometry of the bike and prevent the user from running wider tires and using accessories such as fenders due to the lack of clearance.
## Requirements For Installing 24-inch Wheels On a 20-inch Bicycle
### 1. Clearance
The number one requirement for a 24-inch wheel to fit on a 20-inch bike is clearance.
To learn whether a 24-inch wheel will fit on a 20-inch bike, it’s necessary to do a bit of math.
Below is one of the possible approaches:
1. Measure the radius of the old wheel with the tire pumped to the needed air pressure.
2. Measure the distance between the fork’s crown and the point of the wheel the closest to it.
Combine the radius of the old wheel with the measured distance.
3. Measure the radius of the new wheel with the tire pumped to the needed air pressure.
If the radius of the new wheel isn’t at least 3mm smaller than the combined sum (old radius + fork clearance), then the fork doesn’t have enough clearance to safely accept the 24-inch wheel.
The same procedure can be repeated for the rear wheel, but in that case, the user has to measure two different distances:
a. The distance between the old wheel and the seatstay bridge.
b. The distance between the old wheel and the chainstay bridge.
Of course, in many cases, the user will not have access to a new 24″ inch wheel.
In that case, one can rely on data from a calculator such as this one.
Example:
1. The original 20″ wheel is equipped with a 2″ tire and has a 507.60mm diameter and a 253.8mm radius.
2. The new 24″ wheel is equipped with a 1.5″ tire and has a 583.20mm diameter and a 291.6mm radius.
Or in other words, the new wheel is 37.8mm or 3.78cm/1.48in larger.
In order for the new wheel to fit, there should be at least 4cm/1.57in distance between the fork and the original wheel. If the distance is 3.78cm or less, then the new wheel won’t fit unless the user switches to an even skinnier tire.
In the vast majority of the cases, the forks and frames of 20″ bikes come with a significantly smaller clearance (e.g., 2cm). Thus, in most situations, the user will be unable to install a 24″ wheel on a 20″ bike.
## 2. Brakes
Another important aspect of this conversion is the brake system.
If the bike relies on rim brakes, they won’t be operational upon installing the new wheels because the rim will sit higher in relation to the brake shoes. As a result, the brakes won’t be able to grab the brake strip.
There are solutions to this problem but none of them can be described as super convenient. The options are:
• Adapters elevating the brakes bosses
• Re-welding/re-brazing of the brake bosses
• Installation of clamp-on V-brake mounts
If the bike uses disc brakes, the problem described above does not manifest because the rotor is always located at the center regardless of wheel size. Thus, even when you switch to larger tires, the rotor ends up at the right location.
## 1. Altered geometry
The larger tires will raise the bottom bracket of the bicycle and make it more difficult to take sharp corners due to the higher center of gravity.
If only a front tire is installed, then the head tube angle (HTA) of the bike will get notably slacker. The new head tube angle will change the handling of the bike. The effect will be very noticeable at slow speeds.
Installing a 24″ inch wheel only at the back is dangerous because the tire will steepen the HTA and create a real possibility for the rider to go over the handlebars. Therefore, this choice is both illogical and dangerous.
### 2. Microscopic Tire Clearance
The new wheels will greatly limit the tire clearance of the bike and thus make it impossible to use wide tires and accessories such as full fenders.
Also, if the tire clearance is smaller than 3mm, even a small stone can cause the wheel to jam and trigger an accident.
### 3. Extra Weight
All things being equal, a larger wheel is a heavier wheel. Thus, by switching to 24″ wheels, the rider may make the bike heavier.
### 4. Weaker Wheels
When all parameters are equal, a smaller wheel is a stronger wheel. The shorter radius of smaller wheels reduces the leverage exerted by the rim. Thus, it takes more stress for the wheel to buckle.
That said, a quality 24″ inch wheel is incredibly strong and close to unbreakable unless there’s an accident.
### 5. Slower Acceleration
It’s easier to reach a high RPM (rotation of the cranks per minute) with smaller wheels. Or in other words, larger wheels reduce the bike’s acceleration properties.
However, bigger wheels have the potential to be faster overall. For example, if two bikes have the same high gear, the one with the larger wheels can potentially reach a greater top speed in that gear thanks to the larger circumference.
Or in simpler words, the larger tire covers more ground for the same number of rotations and can therefore be faster.
## FAQ: Does a conversion from 20″ to 24″ wheels have a positive side when the bike is built for 20″ wheels?
Conversion to 24″ wheels can theoretically offer the following benefits:
1. Higher top speed (explained above)
2. Greater roll-over-ability (larger tires can roll over bigger obstacles)
3. A larger bike (bigger tires make smaller bikes feel larger).
However, in practice, the negatives of the conversion make the plan a bit shaky. In most cases, it’s better to sell the existing bike and get a dedicated 24″ or even a 26″ model. | 1,350 | 5,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-38 | latest | en | 0.92629 |
https://www.physicsforums.com/threads/energy-lost-to-friction.223493/ | 1,575,711,056,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540497022.38/warc/CC-MAIN-20191207082632-20191207110632-00286.warc.gz | 833,470,955 | 14,099 | # Energy lost to friction
1. Homework Statement
A spool of thin wire (with inner radius r = 0.40 m, outer radius R = 0.55 m, and moment of inertia I_cm = 0.7139 kg*m^2) pivots on a shaft. The wire is pulled down by a mass M = 1.350 kg. After falling a distance D = 0.550 m, starting from rest, the mass has a speed of v = 65.4 cm/s. Calculate the energy lost to friction during that time.
im not really sure how to do this. i would like any ideas. | 135 | 449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-51 | latest | en | 0.932347 |
https://joyfulparenting.co/2021/02/17/rules-of-the-game/ | 1,679,458,125,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943749.68/warc/CC-MAIN-20230322020215-20230322050215-00315.warc.gz | 408,411,667 | 28,241 | # Rules of the Game
This activity is somewhat of a repeat from when Travis created his own board game in preschool. But now that he’s older, we delved much more deeply not only into how to design a game from start to finish, but also talked about what made a game successful!
To start, we explored two classic games. First up was Dominoes, playing a round with the set we have at home. I had never actually read the real rules before, and when we looked them up online, they were so convoluted I confess even I didn’t quite understand! That was a good jumping off point to talk about what made a game fun and/or challenging.
Next, Travis listened to a read-through of Jumanji, which is a fantastical game of course but a great way to talk about the rules, including what was similar to real-life games, and what was different. (Note: We also watched the movie, but there are scenes that are quite intense and I don’t recommend it for young children).
After all that, it was time to design his own game! Raddish Kids had a lesson plan including an Inventor Inspiration Guide to help kids decide what to base their game around. This involved a ranking system based on likes (food, hobbies, favorite shows or books), but this was all too complicated for my first grader. Travis knew what he wanted to base his game around anyway: Star Wars!
We quickly came up with a game called ‘Race to the Death Star’. The shape of this iconic Star Wars base helped us decide how to configure the game, as a spiral of galaxies closing in on the Dearth Star in the center.
I started gluing down squares of construction paper as the spaces on a large sheet of poster board, and we filled in the ideas as they came to us. Spaces contained events with either a boon (letting the player move forward) or a set-back (which required moving backwards). If a player landed on top of another player, that person had to wait in one of the corner planets until rolling the correct number on the dice.
To make it through “hyperspace” between galaxies required an exact roll. All in all, it actually made for a great board game! Travis loved it so much that we immediately played 3 rounds. He decided on Lego figures as our playing pieces.
As a final component of learning, we explored games that can be played virtually. Travis watched a suggested link of 20 games to play over Zoom, and then we really did Zoom his grandmother to play ‘Zoomed In’ (a game involving close-up images that players take turns guessing). This was such a neat bonding activity and the full video is worth a watch if your kids are currently Zooming relatives and friends.
There was lots more in the lesson plan from Raddish Kids, including suggestions to reinvent an old game with new rules, explore the idea of interactive books like Press Here, or learn the history of a classic board game. But my first grader was gamed out, so we’ll just be here busy playing ‘Race to the Death Star’! | 638 | 2,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-14 | latest | en | 0.984885 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=135418 | 1,371,605,638,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707439012/warc/CC-MAIN-20130516123039-00010-ip-10-60-113-184.ec2.internal.warc.gz | 300,573,776 | 1,181 | ```Question 180501
A business invests \$8,000 in a savings account for two years. At the beginning of the second year, an additional \$2,500 is invested. At the end of the second year, the account balance is \$11,445. What was the annual interest rate?
:
Let x = interest rate in decimal form
:
8000(x+1) = 1st year total
or
8000x + 8000
:
2nd year total
(x+1)(8000 + 8000x + 2500) = 11445
(x+1)(8000x + 10500) = 11445
FOIL
8000x^2 + 10500x + 8000x + 10500 - 11445 = 0
:
8000x^2 + 18500x - 945 = 0
:
simplify divide equation by 5
1600x^2 + 3700x - 189 = 0
:
Use the quadratic formula to solve this:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
:
a=1600; b=3700, c=-189
{{{x = (-3700 +- sqrt(3700^2 - 4 * 1600 * -189 ))/(2*1600) }}}
{{{x = (-3700 +- sqrt(13690000 + 1209600 ))/(3200) }}}
{{{x = (-3700 +- sqrt(14899600 ))/(3200) }}}
Positive solution is what we want here:
{{{x = (-3700 + 3860)/(3200) }}}
{{{x = 160/3200}}}
x = .05, Therefore 5% is the interest rate
;
:
Check solution:
1st year 1.05*8000 = 8400
2nd year 1.05(8400 + 2500) = 11445, confirms our solution
``` | 434 | 1,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2013-20 | latest | en | 0.763172 |
https://fr.scribd.com/document/254253323/ICS-2205-Digital-Logic-printready | 1,571,238,034,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668994.39/warc/CC-MAIN-20191016135759-20191016163259-00433.warc.gz | 500,688,662 | 79,079 | Vous êtes sur la page 1sur 3
KIMATHI UNIVERSITY COLLEGE OF TECHNOLOGY
University Examinations 2010/2011
SECOND YEAR SUPPLEMENTARY/SPECIAL EXAMINATION FOR THE DEGREE OF BACHELOR OF SCIENCE IN COMPUTER SCIENCE/ INFORMATION TECHNOLOGY
ICS 2205: Digital Logic
DATE: 10 th November 2010
TIME: 8.30 am 10.30 am
Instructions: Answer Question 1 and Any Other Two.
INSTRUCTIONS
Answer question one and any other two questions Question one
(a) (i) Explain the main feature of Gray code which is different to ordinary binary code? (2marks) (b) ((i) Drive the Boolean equation for the circuit below (2marks)
(ii) use Boolean algebra to simplify the equation of (b i) above (6marks) (iii) Implement the minimized circuit in (b ii) above using basic gates. (4marks)
Page 1 of 3
• (c) Explain what you understand by the following terms: Standard sum of
(i)
products form (SOP) and Standard product of sums form (POS) (4 marks) (ii) Implement the POS expression (A + B)(B + C + D) (A + C) using
NAND gates only
(4 marks)
• (d) With the aid of digital logic circuits show how the NOR gates can
used to implement the AND, OR and NAND functions
(5 marks)
• (e) Using truth table explains the operation of an EX-OR gate and draw
symbol
(3marks)
Question 2 A mission control systems uses four indication lights A, B, C to monitor the performance of its fuel combustion, it required that a warning buzzer is to sound when the following conditions occur.
• 1. A and B are on
• 2. A, B and C are on
• 3. B, on C are on
• 4. A and C are down
(a)
(i)
Represent the above conditions of the system in a truth table
(3marks)
(ii)
Drive the Boolean equation for output from the resulting truth
table in (a i) above
(4marks)
(b)
Simplify the above Boolean expression
(5marks)
(c)
(i)
Explain the advantages of edge triggered flip-flops over the
level triggered flip-flops.
(ii)
(2 marks)
Describe the operation of D- Latch
(6 marks)
### Question
3
• (a) With the aid of a truth table drive the expression for the sum
(i)
and carry of a full Adder
(4 marks)
(ii)
Reduce the expressions in (i) using Boolean algebra and then
Implement the logic circuit using basic gates
(7 marks)
• (b) the
With
aid
of block diagram
Show how
a
can
be
(6 marks)
• (c) the logic circuit and the truth table of a S-R flip flop using
Sketch
NOR gates
(3marks)
Page 2 of 3
### Question
4
• (a) bit Gray
(i) Design
a three
code
to
binary code converter and
implement the circuit using exclusive-OR gates only (8 marks)
(ii) With
the
aid
of
a block diagram, describe the operation of a
1-of-4 multiplexer. (6marks) (b) State the applications of following digital logic circuits: (i) Encoders (ii) Decoders (iii) Ring counter (3 marks) (c) State limitations of the ripple through (asynchronous) counters. (3 marks)
### Question
5
• (a) Use Boolean algebra proof that
(i)
(3marks)
(ii)
Convert the following binary numbers into gray code and from
gray to binary
(2marks)
Binary numbers
Gray code
1101111
1001101
1010101
1111001
• (b) With aid of an appropriate block and timing diagrams diagram s
describe how 4 bit serial shift register can be loaded with the
binary data 1011
(8marks)
• (c) Use K-map or otherwise to simplify the logic diagram below.
(i)
(5marks) (ii) Draw simplified logic diagram (2 marks)
Page 3 of 3 | 920 | 3,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-43 | latest | en | 0.773275 |
https://www.theochem.ru.nl/~pwormer/Knowino/knowino.org/wiki/Free_space_(electromagnetism).html | 1,686,422,855,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657735.85/warc/CC-MAIN-20230610164417-20230610194417-00115.warc.gz | 1,146,764,453 | 12,269 | # Free space (electromagnetism)
Free space usually refers to a perfect vacuum, devoid of all particles. The term is most often used in classical electromagnetism where it refers to a reference state sometimes called classical vacuum,[1] and in quantum physics where it refers to the ground state of the electromagnetic field, which is subject to fluctuations about a dormant zero average-field condition.[2] The classical case of vanishing fields implies all fields are source-attributed, while in the quantum case field moments can arise without sources by virtual photon creation and destruction.[3] The description of free space varies somewhat among authors, with some authors requiring only the absence of substances with electrical properties,[4] or of charged matter (ions and electrons, for example).[5]
## Contents
### Classical case
In classical physics, free space is a concept of electromagnetic theory, corresponding to a theoretically perfect vacuum and sometimes referred to as the vacuum of free space, or as classical vacuum, and is appropriately viewed as a reference medium.[1] In the classical case, free space is characterized by the electrical permittivity ε0 and the magnetic permeability μ0.[6] The exact value of ε0 is provided by NIST as the electric constant [7] and the defined value of μ0 as the magnetic constant:[8]
ε0 ≈ 8.854 187 817... × 10−12 F m−1
μ0 = 4π × 10−7 ≈ 12.566 370 614... x 10−7 N A−2
where the approximation is not a physical uncertainty (such as a measurement error) but a result of the inability to express these irrational numbers with a finite number of digits.
One consequence of these electromagnetic properties coupled with Maxwell's equations is that the speed of light in free space is related to ε0 and μ0 via the relation:[9]
$c_0 = 1/\sqrt{\mu_0 \varepsilon_0}\ .$
Using the defined valued for the speed of light provided by NIST as:[10]
c0 = 299 792 458 m s −1,
and the already mentioned defined value for μ0, this relationship leads to the exact value given above for ε0.
Another consequence of these electromagnetic properties is that the ratio of electric to magnetic field strengths in an electromagnetic wave propagating in free space is an exact value provided by NIST as the characteristic impedance of vacuum:[11]
$Z_0 = \sqrt{\mu_0 /\varepsilon_0} \$
= 376.730 313 461... Ω.
It also can be noted that the electrical permittivity ε0 and the magnetic permeability μ0 do not depend upon direction, field strength, polarization, or frequency. Consequently, free space is isotropic, linear, non-dichroic, and dispersion free. Linearity, in particular, implies that the fields and/or potentials due to an assembly of charges is simply the addition of the fields/potentials due to each charge separately (that is, the principle of superposition applies).[12]
### Quantum case
For a discussion of the quantum case, see Vacuum (quantum electrodynamic).
### Attainability
A perfect vacuum is itself only realizable in principle.[13][14] It is an idealization, like absolute zero for temperature, that can be approached, but never actually realized:[13]
“One reason [a vacuum is not empty] is that the walls of a vacuum chamber emit light in the form of black-body radiation...If this soup of photons is in thermodynamic equilibrium with the walls, it can be said to have a particular temperature, as well as a pressure. Another reason that perfect vacuum is impossible is the Heisenberg uncertainty principle which states that no particles can ever have an exact position...More fundamentally, quantum mechanics predicts ... a correction to the energy called the zero-point energy [that] consists of energies of virtual particles that have a brief existence. This is called vacuum fluctuation.” Luciano Boi, "Creating the physical world ex nihilo?" p. 55
And classical vacuum is one step further removed from attainability because its permittivity ε0 and permeability μ0 do not allow for quantum fluctuations. Nonetheless, outer space and good terrestrial vacuums are modeled adequately by classical vacuum for many purposes.
## References
1. 1.0 1.1 Werner S. Weiglhofer and Akhlesh Lakhtakia (2003). “§4.1: The classical vacuum as reference medium”, Introduction to complex mediums for optics and electromagnetics. SPIE Press. ISBN 0819449474.
2. Ramamurti Shankar (1994). Principles of quantum mechanics, 2nd ed.. Springer, p. 507. ISBN 0306447908.
3. Werner Vogel, Dirk-Gunnar Welsch (2006). Quantum optics, 3rd ed.. Wiley-VCH, p. 337. ISBN 3527405070.
4. RK Pathria (2003). The Theory of Relativity, Reprint of Hindustan 1974 2nd ed.. Courier Dover Publications, p. 119. ISBN 0486428192.
5. (1992) Christopher G. Morris, editor: Academic Press dictionary of science and technology. Academic, p. 880. ISBN 0122004000.
6. Akhlesh Lakhtakia, R. Messier (2005). “§6.2: Constitutive relations”, Sculptured thin films: nanoengineered morphology and optics. SPIE Press, p. 105. ISBN 0819456063.
7. CODATA. Electric constant. 2006 CODATA recommended values. NIST. Retrieved on 2010-11-28.
8. CODATA. Magnetic constant. 2006 CODATA recommended values. NIST. Retrieved on 2010-11-28.
9. Albrecht Unsöld, B. Baschek (2001). “§4.1: Electromagnetic radiation, Equation 4.3”, The new cosmos: an introduction to astronomy and astrophysics, 5th ed.. Springer, p. 101. ISBN 3540678778.
10. CODATA. Speed of light in vacuum. 2006 CODATA recommended values. NIST. Retrieved on 2010-11-28. A defined value for the speed of light is a consequence of adoption of time of transit as the measure of length, so lengths are measured in seconds. See metre.
11. CODATA. Characteristic impedance of vacuum Z0. 2006 CODATA recommended values. NIST. Retrieved on 2010-11-28.
12. A. Pramanik (2004). “§1.3 The principle of superposition”, Electro-Magnetism: Theory and Applications. PHI Learning Pvt. Ltd, pp. 37-38. ISBN 8120319575.
13. 13.0 13.1 Luciano Boi (2009). “Creating the physical world ex nihilo? On the quantum vacuum and its fluctuations”, Ernesto Carafoli, Gian Antonio Danieli, Giuseppe O. Longo, editors: The Two Cultures: Shared Problems. Springer, p. 55. ISBN 8847008689.
14. PAM Dirac (2001). Jong-Ping Hsu, Yuanzhong Zhang, editors: Lorentz and Poincaré invariance: 100 years of relativity. World Scientific, p. 440. ISBN 9810247214. | 1,585 | 6,308 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-23 | latest | en | 0.901063 |
https://www.transtutors.com/questions/a-computer-program-is-tested-by-3-independent-tests-when-there-is-an-error-these-tes-3951472.htm | 1,582,327,180,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145538.32/warc/CC-MAIN-20200221203000-20200221233000-00452.warc.gz | 923,427,448 | 15,942 | # A computer program is tested by 3 independent tests. When there is an error, these tests will... 1 answer below »
A computer program is tested by 3 independent tests. When there is an error, thesetests will discover it with probabilities 0.2, 0.3, and 0.5, respectively. Suppose that the program contains anerror. What is the probability that it will be found by at least one test?
k n
Here A computer program is tested by 3 independent tests
Given probabilities 0.2, 0.3 and 0.5
Let A = error found by test 1 | 136 | 517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-10 | latest | en | 0.954453 |
http://www.gradesaver.com/textbooks/math/prealgebra/prealgebra-7th-edition/chapter-8-cumulative-review-page-601/38 | 1,524,354,513,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945459.17/warc/CC-MAIN-20180421223015-20180422003015-00251.warc.gz | 425,616,174 | 13,359 | ## Prealgebra (7th Edition)
To convert a percentage to a decimal, move the decimal point 2 places to the left. $\overset\curvearrowleft{002}.7\%$ | 43 | 146 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-17 | latest | en | 0.604948 |
https://documen.tv/question/a-mater-eam-q2-sa-1-4-points-what-is-the-slope-of-a-line-that-is-perpendicular-to-the-line-f-2-6-21487834-45/ | 1,675,051,774,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499801.40/warc/CC-MAIN-20230130034805-20230130064805-00770.warc.gz | 236,528,379 | 19,941 | ## A Mater Exam Q2 SA 1 4 Points What is the slope of a line that is perpendicular to the line f(x) = 2x + 6? Justify your an
Question
A Mater Exam
Q2 SA 1
4 Points
What is the slope of a line that is perpendicular to the line f(x) = 2x + 6? Justify your answer
Select file(s)
in progress 0
1 year 2021-08-29T20:39:08+00:00 1 Answers 0 views 0
The slope of the perpendicular line to the line is
m = $$\frac{-1}{2}$$
Step-by-step explanation:
Step(i):-
Given that the line y =f(x) = 2x +6
⇒ y = 2x +6
comparing y =mx +c
The slope of the given line m =2
Step(ii):-
The slope of the perpendicular line = $$\frac{-1}{m}$$
The slope of the perpendicular line = $$\frac{-1}{2}$$ | 243 | 707 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-06 | latest | en | 0.766862 |
https://blog.ipspace.net/2013/05/data-has-mass-and-gravity.html | 1,718,436,237,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861584.65/warc/CC-MAIN-20240615062230-20240615092230-00168.warc.gz | 117,925,941 | 20,527 | ## Data Has Mass and Gravity
A while ago, while listening to an interesting CloudCast podcast (my second favorite podcast - the best one out there is still the Packet Pushers), I stumbled upon an interesting idea “Data has gravity”. The podcast guest used that idea to explain how data agglomerates in larger and larger chunks and how it makes sense to move the data processing (application) closer to the data.
Look up the data gravity formula on datagravity.org. It’s almost identical to Newton’s law of universal gravitation, with latency adjusted for bandwidth playing the part of distance.
Having spent the last few months as an occasional “physics consultant” for my high-school daughter, I loved the idea, quickly mentally transformed it to “Data has mass” (which then generates gravity) and immediately started drawing additional parallels that you should keep in mind every time someone starts blabbering about long-distance VM migration or cloudbursting.
Cloudbursting (noun) - the "visionary" idea of dynamically moving workloads into the cloud, while forgetting the dirty details like IP routing, access to data, latency, bandwidth and a few other minor details. Works best in the reality-distortion field of PowerPoint presentations.
Data is hard to move. The energy needed to move an object is proportional to object's mass and the speed with which you want to move the object. In networking terms: you need more bandwidth (more energy) or more time (because you're moving data at lower speed). The follow-up CloudCast podcast has a great example of someone who spent four days moving the data so he could run a 15-minute computation on it.
Data creates gravity well. You have to reach pretty high speed (and spend a significant amount of energy) if you want to escape from a large mass. In application terms: Once an application (and its users) becomes accustomed to low latency access to data, it takes an enormous amount of energy to move that application somewhere else, sometimes requiring a total rewrite of the application (see also: going from Cessna to Space Shuttle).
Here’s the more formal explanation of gravity well. As is usually the case, xkcd provides a more down-to-Earth explanation.
A similar phenomenon (although on a smaller scale, because vMotion limits latency to 10msec) happens when you vMotion a live VM away from its data. In the famous white paper where Cisco, EMC and VMware managed to move a running SQL server into another data center, the performance dropped by ~30% until they made the LUN closer to the VM active. If you ever had the "privilege" of talking with someone over a satellite connection, you know how that poor SQL server felt.
In more humorous terms, the further away from the data the application is, the lower the data’s gravity force is … until the application gets (v)motion sickness from lack of gravity ;)
Data attracts more data. An obvious corollary to the previous two facts: as you don't move the data (because it's too bulky) and the applications using the data live close to it (to enjoy the low-latency environment), those applications generate even more data. The bulky data heap only grows, generating an even higher gravity well.
Disclaimer: I'm not saying that "moving workloads into the cloud" is hogwash. What I'm saying is that a move to the cloud must be a conscious, well-planned decision and that you have to figure out what to do with the data before moving the compute workload … a fact that is well obscured by point-and-click eye candy offered by several vendors (you know, you just click on a VM … and **unicorn magic** it runs in the cloud). | 761 | 3,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-26 | latest | en | 0.922201 |
http://stackoverflow.com/questions/10987296/breadth-first-search-depth-first-search-or-directed-graph | 1,464,681,379,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464051196108.86/warc/CC-MAIN-20160524005316-00111-ip-10-185-217-139.ec2.internal.warc.gz | 284,606,035 | 22,117 | # breadth first search / depth first search or Directed graph?
I have been struggling with this for a while now. given a set of nodes:
``````nodes = { ('A','B'),
('B','C'),
('C','D'),
('C','E'),
('B','E'),
('C','F') }
``````
what is the best way to achieve the following:
`````` A
|
B
_________|_________
| |
C E
_____|_____ |
| | | C
D E F ____|____
| |
D F
``````
where I can see that:
``````the routes from A -> B:
A -> B
the routes from A -> C:
A -> B -> C
A -> B -> E -> C
the routes from A -> D:
A -> B -> C -> D
A -> B -> E -> C -> D
etc...
``````
My reason for doing this, is purely because I want to understand how to.
I know that bfs finds the quickest route, (I think I might be using something similar in the get children function)
but I do not know the best way to loop / recursively run over the graph. Should I use a dictionary and work with key/vals or a list. Or sets...
``````def make_graph(nodes):
d = dict()
for (x,y,*z) in nodes:
if x not in d: d[x] = set()
if y not in d: d[y] = set()
return d
``````
I am using *z here as the tuples will actually include a float, but at the moment I am trying to keep things simple.
``````def display_graph(nodes):
for (key,val) in make_graph(nodes).items():
print(key, val)
# A {'B'}
# C {'B', 'E', 'D', 'F'}
# B {'A', 'C', 'E'}
# E {'C', 'B'}
# D {'C'}
# F {'C'}
``````
the getchildren function finds all possible end points for the node root:
``````def getchildren(noderoot,graph):
previousnodes, nextnodes = set(), set()
currentnode = noderoot
while True:
nextnodes.update(graph[currentnode] - previousnodes)
try:
currentnode = nextnodes.pop()
except KeyError: break
return (noderoot, previousnodes - set(noderoot))
``````
In this case A:
``````print(getchildren('A', make_graph(nodes)))
# ('A', {'C', 'B', 'E', 'D', 'F'})
``````
-
possible duplicate of How can I convert a series of parent-child relationships into a hierarchical tree?. Not the same language, but same problem – Eric Jun 11 '12 at 20:49
Why does `E` not appear under `C` on the left? – Eric Jun 11 '12 at 20:52
Two each of C, D, F? Are you sure you want a tree, and not a directed graph? – Hugh Bothwell Jun 11 '12 at 20:52
I want to find all possible routes for any given key. I am not searching for shortest path ... and I do not know my 'goal'. I don't know if I want a tree or a graph, most of this was written without any prior knowledge of how trees and graphs actually work – beoliver Jun 11 '12 at 20:54
@Eric I'm not sure where you mean? – beoliver Jun 11 '12 at 20:59
Before coding with a program language, you need to abstract the problem properly.
First you need to think about the properties of your graph, such as cyclic/acyclic, directed/undirected, etc..
Then you need to choose a way to solve your problem accordingly. e.g. if it's a acyclic, undirected and connected graph, then you can represent the graph as a tree and use either BFS or DFS to traverse it.
Finally, after you think through all of these, you can put it into code much more easier. Like what you've already been doing, you can give each node a list storing all the neighbors and use the BFS to traverse the tree.
-
I'n my head, I know how to do it, but when I try and code it... I get confused. – beoliver Jun 11 '12 at 21:40
@ThemanontheClaphamomnibus put what's in your mind into nature language --> put nature language into pseudo code --> put pseudo code into real code – xvatar Jun 11 '12 at 21:50
Thank you everyone, Problem solved. The function I needed to write was the following.
``````def trace_graph(k, graph):
""" takes a graph and returns a list of lists showing all possible routes from k """
paths = [[k,v] for v in graph[k]]
for path in paths:
xs = path[:-1]
x = path[-1]
for v in graph[x]:
if v not in xs and path + [v] not in paths:
paths.append(path + [v])
paths.sort()
return paths
for path in trace_graph('A', make_graph(nodes)):
print(path)
['A', 'B']
['A', 'B', 'C']
['A', 'B', 'C', 'D']
['A', 'B', 'C', 'E']
['A', 'B', 'C', 'F']
['A', 'B', 'E']
['A', 'B', 'E', 'C']
['A', 'B', 'E', 'C', 'D']
['A', 'B', 'E', 'C', 'F']
``````
-
I don't think an ordinary tree structure makes sense for representing your data, given that it's sequential but not necessarily ordered/sorted. It would probably be more appropriate to make use of either tries (either prefix or radix trees) or (probably better) directed graphs.
-
I think you may be making things more complicated than they need to be. Think about the type of data you are representing as xvatar has said.
For a basic directed graph a dictionary makes sense. Simply store parent: list of children.
``````nodes = [ ('A','B'),
('B','C'),
('C','D'),
('C','E'),
('B','E'),
('C','F') ]
from collections import defaultdict
d = defaultdict(list)
for node in nodes:
d[node[0]].append(node[1])
``````
Finding all reachable children from any root node is simple:
``````def getchildren(root, graph, path=[]):
path = path + [root]
for child in graph[root]:
if child not in path: #accounts for cycles
path=getchildren(child, graph, path)
return path
``````
When calling:
``````>>> print getchildren('A',d)
['A', 'B', 'C', 'D', 'E', 'F']
>>> print getchildren('C',d)
['C', 'D', 'E', 'F']
``````
-
But surely I can do this at the moment, the getchildren function does exactly this It just used sets instead of lists. in your getchildren function, why does C not return A - as (A,B),(C,B) – beoliver Jun 11 '12 at 21:51
Ah... In my mind (A,B) == (B,A) - where they are undirected – beoliver Jun 11 '12 at 21:57
@ThemanontheClaphamomnibus I assumed directed... as in the question title... You asked for the best way to recursively run over the graph and to store data and I recommended using a defaultdict for easy creation and a simple recursive algorithm that will handle cycles when traversing. If you're already doing what you wanted, what are you really asking? – BasilV Jun 11 '12 at 23:00 | 1,713 | 6,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2016-22 | latest | en | 0.812396 |
https://www.geeksforgeeks.org/optimal-substructure-property-in-dynamic-programming-dp-2/amp/?ref=rp | 1,596,805,736,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737178.6/warc/CC-MAIN-20200807113613-20200807143613-00396.warc.gz | 685,577,925 | 16,688 | # Optimal Substructure Property in Dynamic Programming | DP-2
As we discussed in Set 1, following are the two main properties of a problem that suggest that the given problem can be solved using Dynamic programming:
1) Overlapping Subproblems
2) Optimal Substructure
We have already discussed Overlapping Subproblem property in the Set 1. Let us discuss Optimal Substructure property here.
2) Optimal Substructure: A given problems has Optimal Substructure Property if optimal solution of the given problem can be obtained by using optimal solutions of its subproblems.
For example, the Shortest Path problem has following optimal substructure property:
If a node x lies in the shortest path from a source node u to destination node v then the shortest path from u to v is combination of shortest path from u to x and shortest path from x to v. The standard All Pair Shortest Path algorithms like Floyd–Warshall and Bellman–Ford are typical examples of Dynamic Programming.
On the other hand, the Longest Path problem doesn’t have the Optimal Substructure property. Here by Longest Path we mean longest simple path (path without cycle) between two nodes. Consider the following unweighted graph given in the CLRS book. There are two longest paths from q to t: q→r→t and q→s→t. Unlike shortest paths, these longest paths do not have the optimal substructure property. For example, the longest path q→r→t is not a combination of longest path from q to r and longest path from r to t, because the longest path from q to r is q→s→t→r and the longest path from r to t is r→q→s→t.
We will be covering some example problems in future posts on Dynamic Programming. | 369 | 1,662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-34 | latest | en | 0.910322 |
https://kr.mathworks.com/matlabcentral/answers/195034-matlab-analytical-ft-and-fft-comparison | 1,660,239,062,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571483.70/warc/CC-MAIN-20220811164257-20220811194257-00339.warc.gz | 336,647,173 | 27,792 | # Matlab analytical FT and FFT comparison
조회 수: 22(최근 30일)
Ongun 2015년 3월 28일
편집: Youssef Khmou 2015년 3월 29일
My aim is to compare the analytical Fourier transform of a function with the FFT of the same function using Matlab routines and correct normalization coefficients. In order to evaluate the analytical FT I used the function fourier and for FFT I use the function fft while normalizing with respect to the length of the vector. Later I plot the FFT till the Nyquist frequency and also plot the results of the analytical FT on the same plot but their centers and height does not match. I am not that familiar with FFTs and I would really appreciate it if someone pointed me in the right direction. You can check the working code snippet below:
close all;
clear all;
clc;
N = 100;
dt = 0.01;
Fs = 1/dt;
F = zeros(N,1);
n0 = 50;
for n = 1:N
F(n) = exp(-((n-n0)*dt)^2*pi^2);
end
G_cmp = fft(F)/N;
f_arr = Fs/2*linspace(0,1,N/2+1);
syms t f
F_an = exp(-(t-n0*dt)^2*pi^2);
G_an = fourier(F_an, t, f);
f1 = figure;
p1 = plot(dt*(1:N), F);
saveas(f1, 'pulse_time.png');
f2 = figure;
p2 = plot(f_arr, 2*abs(G_cmp(1:N/2+1)), f_arr, 2*abs(subs(G_an, 'f', f_arr)));
saveas(f2, 'pulse_freq.png');
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### 답변(1개)
Youssef Khmou 2015년 3월 29일
편집: Youssef Khmou 2015년 3월 29일
numerical fft requires a shift if we want to visualize the spectrum with both negative and positive frequencies, scaling problem is not yet solved, however try the following version, the theoretical transformation is calculated using 2*pi*f instead of f :
close all;
clear all;
clc;
N = 100;
dt = 0.01;
Fs = 1/dt;
F = zeros(N,1);
n0 = 50;
for n = 1:N
F(n) = exp(-((n-n0)*dt)^2*(pi^2));
end
G_cmp = fft(F)/((N));
f_arr = Fs/2*linspace(-1,1,N);
syms t f
F_an = exp(-(t-n0*dt)^2*(pi^2));
G_an = fourier(F_an, t, 2*pi*f);
f1 = figure;
subplot(1,2,1)
p1 = plot(dt*(1:N), F);
%saveas(f1, 'pulse_time.png');
%f2 = figure;
subplot(1,2,2)
p2 = plot(f_arr, fftshift(2*abs(G_cmp)),'-+', f_arr, 2*abs(subs(G_an, 'f', f_arr)),'r--');
%saveas(f2, 'pulse_freq.png');
##### 댓글 수: 3표시숨기기 이전 댓글 수: 2
Youssef Khmou 2015년 3월 29일
you are welcome, i have written an fft code previously, here is the implementation :
%==========================================================================
% function z=Fast_Fourier_Transform(x,nfft)
%
% N=length(x);
% z=zeros(1,nfft);
% Sum=0;
% for k=1:nfft
% for jj=1:N
% Sum=Sum+x(jj)*exp(-2*pi*j*(jj-1)*(k-1)/nfft);
% end
% z(k)=Sum;
% Sum=0;% Reset
% end
however for large vector, it is time consuming.
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mathwithpython.com | 1,386,754,416,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164034245/warc/CC-MAIN-20131204133354-00081-ip-10-33-133-15.ec2.internal.warc.gz | 117,919,447 | 54,450 | # // <![CDATA[</p> <p>// ]]>What happens when we toss a dice with more than 6 sides?// <![CDATA[ init_mathjax = function() { if (window.MathJax) { // MathJax loaded MathJax.Hub.Config({ tex2jax: { inlineMath: [ ['$','$'], ["\$","\$"] ], displayMath: [ ['$$','$$'], ["\$","\$"] ] }, displayAlign: 'left', // Change this to 'center' to center equations. "HTML-CSS": { styles: {'.MathJax_Display': {"margin": 0}} } }); MathJax.Hub.Queue(["Typeset",MathJax.Hub]); } } init_mathjax(); // ]]>
In a previous post we looked at the question:
How many times, on average do you need to toss a dice with 6 sides until you get a 6?
Python has some cool randomness features so let’s look at those.
In [1]:
import random
In [2]:
random.choice?
random.choice is pretty cool. Here’s the documentation:
Choose a random element from a non-empty sequence.
We can give it a list of stuff and it’ll pick on out at random.
Here’s a dice with 6 sides:
This is easy to create with Python
In [3]:
sides = [1, 2, 3, 4, 5, 6]
In [4]:
random.choice(sides)
Out[4]:
4
In [5]:
random.choice(sides)
Out[5]:
1
Every time we call random.choice it randomly pulls a number from our list. Very cool! It’s like pulling a number from a hat.
Pulling random numbers happens quite a lot. In fact, it’s how the NBA drafts new playes fon the upcoming year.
Perhaps I’ll talk about the NBA draft at a later date, let’s continue with our dice example.
random.choice is cool because it allows us to construct any type of dice that want.
Here are some examples
In [6]:
six_sided = [1, 2, 3, 4, 5, 6]
ten_sided_norm = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
ten_sided_irregular = [1, 1, 1, 2, 2, 3, 3, 4, 5, 6]
Previously we had a function that threw the die until we got a certain number.
Let’s write it again, this time using random.choice
In [7]:
def toss_till_value(value, sides):
tosses = []
while 1:
outcome = random.choice(sides)
tosses.append(outcome)
if outcome == value:
break
return tosses
Now we can toss any die we want and get back the results!
We know a little bit about a die with 6 sides, so let’s try that first, tossing it until we get a 4.
In [9]:
toss_till_value(4, six_sided)
Out[9]:
[3, 3, 5, 4]
The result from our function is the hisory of tosses.
Toss it a few more times for fun!
In [11]:
toss_till_value(4, six_sided)
Out[11]:
[1, 6, 4]
In [12]:
toss_till_value(4, six_sided)
Out[12]:
[2, 6, 2, 4]
In [13]:
toss_till_value(4, six_sided)
Out[13]:
[1, 1, 5, 2, 5, 1, 6, 5, 1, 1, 3, 5, 5, 3, 4]
In [14]:
toss_till_value(4, six_sided)
Out[14]:
[6, 1, 6, 3, 2, 3, 5, 5, 5, 6, 2, 5, 6, 5, 1, 3, 3, 3, 5, 6, 2, 1, 4]
Wow! That last resulrt was really long!
Now, let’s use our 10 sided die, stopping when we get a 4.
In [16]:
toss_till_value(4, ten_sided_norm)
Out[16]:
[1, 2, 1, 10, 3, 7, 5, 4]
In [17]:
toss_till_value(4, ten_sided_norm)
Out[17]:
[8, 4]
In [18]:
toss_till_value(4, ten_sided_norm)
Out[18]:
[4]
In [19]:
toss_till_value(4, ten_sided_norm)
Out[19]:
[2, 5, 5, 10, 2, 1, 2, 5, 9, 9, 3, 3, 9, 3, 10, 1, 3, 8, 4]
In [22]:
toss_till_value(4, ten_sided_norm)
Out[22]:
[7, 3, 10, 5, 6, 6, 10, 2, 1, 9, 7, 8, 8, 7, 9, 9, 4]
The previous question was:
How many times, on average, do we have to throw a 6 sided die untill we get a 6?
Now we have a 10 sided dice, so let’s use that?
How many times, on average, do we have to throw a 10 sided die until we get a 6?
What are we going to do here? Can you write it down?
1. Toss the die until we get a 6.
2. Count the number of times we tossed it, and write it down.
3. Repeat step 1 and 2 100 times, untill we have 100 records of the number of throws it took to get a 6.
How do we record, or write down, our numbers? Python has a list! We .append items to our list, which is like writing them down.
In [24]:
number_of_tosses = []
for i in range(100):
result = toss_till_value(6, ten_sided_norm)
# Count how many times we threw the die, in python terms this is called the 'length'
number_of_tosses.append(len(result))
print number_of_tosses
[3, 8, 3, 8, 8, 11, 5, 1, 15, 1, 6, 6, 4, 1, 5, 15, 9, 6, 1, 16, 9, 2, 8, 1, 2, 1, 13, 2, 2, 2, 7, 43, 7, 3, 11, 9, 11, 4, 18, 6, 2, 6, 7, 34, 15, 5, 8, 3, 4, 14, 17, 10, 31, 3, 10, 2, 4, 2, 17, 9, 9, 12, 15, 14, 10, 30, 5, 1, 2, 4, 16, 7, 15, 13, 19, 28, 4, 2, 7, 39, 4, 10, 2, 4, 8, 6, 6, 5, 8, 7, 5, 5, 8, 1, 9, 4, 10, 56, 7, 9]
We can see the results. In snome cases it took 1 throw to get a 6, other times it took as many as 39, 43, and even 56 throws!
To get the average number of throws, we add up all our numbers, and divide by the number of “trails”, in this case it was 100.
In [25]:
sum(number_of_tosses) / 100.
Out[25]:
9.22
This means that it took 9.22 throws on average to get a 6.
What would happen if we did it with 500 throws instead of 100?
I’m going to copy and paste the code above and change 100 -> 500, then re-run our experiment.
In [28]:
number_of_tosses = []
for i in range(500):
result = toss_till_value(6, ten_sided_norm)
# Count how many times we threw the die, in python terms this is called the 'length'
number_of_tosses.append(len(result))
sum(number_of_tosses) / 500.
Out[28]:
9.836
Humm… 9.836. That’s similar to before, but this time a little closer to 10.
What would happen if we did it with 1000, 10000, or even 100000?
Here something nice about doing math with computers, we don’t have to copy and past and do a lot by hand, the computer can do the work for us. But we need to tell the computer how to run our expirement.
This is the same code we used before, I’m just replacing the specific numbers with a place holder.
• 500 -> trials
• 6 -> value
• ten_sided_norm -> dice_type This is special because now we can use any type of dice we want
In [32]:
def run_experiment(value, dice_type, trials):
number_of_tosses = []
for i in range(trials):
result = toss_till_value(value, dice_type)
# Count how many times we threw the die, in python terms this is called the 'length'
number_of_tosses.append(len(result))
return sum(number_of_tosses) / float(trials)
In [33]:
run_experiment(6, ten_sided_norm, 1000)
Out[33]:
10.072
It took, on average, about 10 throws to get a 6 when using a 10 sided die.
To be sure that the average is close to 10, let’s throw it 100000 times (one hundred thousand times!). What do you think will happen?
In [34]:
run_experiment(6, ten_sided_norm, 100000)
Out[34]:
10.02647
Yep, 10!
But I wonder if it changes if we use something other than 6? Let’s try all the numbers!
In [35]:
print ten_sided_norm
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
In [36]:
print " 1:", run_experiment(1, ten_sided_norm, 10000)
print " 2:", run_experiment(2, ten_sided_norm, 10000)
print " 3:", run_experiment(3, ten_sided_norm, 10000)
print " 4:", run_experiment(4, ten_sided_norm, 10000)
print " 5:", run_experiment(5, ten_sided_norm, 10000)
print " 6:", run_experiment(6, ten_sided_norm, 10000)
print " 7:", run_experiment(7, ten_sided_norm, 10000)
print " 8:", run_experiment(8, ten_sided_norm, 10000)
print " 9:", run_experiment(9, ten_sided_norm, 10000)
print "10:", run_experiment(10, ten_sided_norm, 10000)
1: 10.1279
2: 9.9722
3: 9.9648
4: 9.7546
5: 9.9235
6: 10.142
7: 9.98
8: 10.0858
9: 10.0105
10: 10.0697
Wow, all the numbers are pretty close to 10.
I guess it doesn’t matter what face value we choose.
But, I wonder what happens if we start with a 6 sided dice, and increasing the side each time so that we get try a 7 , then an 8, then a 9…
In [37]:
print " 6 sides:", run_experiment(6, [1, 2, 3, 4, 5, 6], 10000)
print " 7 sides:", run_experiment(6, [1, 2, 3, 4, 5, 6, 7], 10000)
print " 8 sides:", run_experiment(6, [1, 2, 3, 4, 5, 6, 7, 8], 10000)
print " 9 sides:", run_experiment(6, [1, 2, 3, 4, 5, 6, 7, 8, 9], 10000)
print "10 sides:", run_experiment(6, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 10000)
print "11 sides:", run_experiment(6, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], 10000)
print "12 sides:", run_experiment(6, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], 10000)
6 sides: 6.0582
7 sides: 6.9162
8 sides: 8.0179
9 sides: 9.1095
10 sides: 9.9055
11 sides: 10.7736
12 sides: 12.1072
What’s going on here?
We have a 6 sided die and it takes 6 throws?
A 7 sided die and it takes 7 throws?
It seams that the number of sides on the dice, is the number of throws, on average it takes.
Let’s test out our idea. How about we try a 50 sided die!
Believe it or not, a 50 sided die actually exists!
In [40]:
fifty_sided_die = [1, 2, 3, 4, 5, 6, 7, 8, 9, \
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, \
20, 21, 22, 23, 24, 25, 26, 27, 28, 29, \
30, 31, 32, 33, 34, 35, 36, 37, 38, 39, \
40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50]
In [42]:
print "50 sides:", run_experiment(6, fifty_sided_die, 10000)
50 sides: 49.8958
That’s pretty close to 50!
The number of sides on the dice is the number of times we have to roll the dice, on average, until we get a 6
A 100-sided die exists too!
Since the computer allowed us to try a 50-sided die without breaking a sweat, let’s do a 100 sided die!
(And you can get it on Amazon.)
Note: not only do I hate calculating, that’s why I want to use a computer, I hate typing stuff out. I don’t want to type out all those numbers.
hundred_sided_die = [1, 2, 3, 4, .... 100]
That’s a lot of typing!
Python has a trick for generating number and it’s:
range
In [41]:
fifty_sided_die == range(1, 51)
Out[41]:
True
We can make any number of sides with a simple range
range(1, sides + 1)
In [44]:
hundred_sided_die = range(1, 101)
In [45]:
print "100 sides:", run_experiment(6, hundred_sided_die, 10000)
100 sides: 100.0524
# Problem
## On average how many throws does it take to get a 6?
People are familiar with a dice that has 6 sides, with each face having 1-6 on each side.
This problem begs for a simulation. Python has a cool random funciton that let’s us simulate a dice throw.
We begin by creating a function that toss 6-sided die until we get a 6.
In [157]:
import random
In [158]:
def toss_till_6():
tosses = []
while True:
face = random.randint(1, 6)
tosses.append(face)
if face == 6:
break
return tosses
We can toss the die
In [164]:
toss_till_6()
Out[164]:
[4, 4, 1, 6]
Let’s toss it 10 times and see what happens
In [162]:
for i in range(10):
print toss_till_6()
[5, 6]
[3, 5, 3, 5, 5, 3, 4, 6]
[3, 2, 5, 2, 2, 2, 2, 4, 5, 3, 6]
[6]
[3, 4, 1, 6]
[2, 3, 2, 6]
[6]
[4, 3, 5, 6]
[5, 3, 2, 1, 5, 6]
[2, 2, 3, 3, 1, 1, 6]
That’s interesting, sometimes we get 6 on the first throw, sometimes it takes quite a while. Check out that big string of 2′s in the third trial.
So, on average, how many throws did it take to get a 6?
Let’s count them up.
• The 1st trial took 2 throws.
• The 2nd trial took 8 throws.
• The 3rd trial took 11 throws.
• The 4th trial took 1 throw.
• The 5th trial took 4 throws.
• The 6th trial took 4 throws.
• The 7th trial took 1 throws.
• The 8th trial took 4 throws.
• The 9th trial took 6 throws.
• The 10th trial took 7 throws.
That’s a lot of calculating. Python can handle the heavy lifting.
In [165]:
throw_counts = [2, 8, 11, 1, 4, 4, 1, 4, 6, 7]
sum(throw_counts)
Out[165]:
48
We threw the dice until we got a 6. And we did it 10 times. It too a total of 48 throws. How many is that on average?
In [168]:
sum(throw_counts) / 10. #Note, that period after the 10 is really important. If we don't use it we won't get a decimal.
Out[168]:
4.8
It took almost 5 throws before we got a 6.
Ah, but sometimes we got a 6 on the first throw, sometimes it took as many as 11 throws.
If we did our experiment one time and got 6 onthe first throw we’d be way off. We’d also be way off if we had the throw the dice 11 times beore getting a 6.
What would happen if we repeated this experiment more times?
How many more… 100, 1000, 10000?
Let’s begin with 100.
(Note: I’m only choosing 100 because it’s more than 10. We could also choose some other number like 82 if we wanted.)
In [169]:
trials = 100
results = []
for i in range(trials):
trial = toss_till_6()
results.append(trial)
results
Out[169]:
[[2, 3, 3, 3, 3, 3, 2, 5, 5, 1, 2, 4, 6],
[5, 4, 4, 4, 6],
[2, 6],
[6],
[3, 4, 4, 3, 3, 4, 4, 6],
[4, 5, 1, 4, 2, 5, 1, 1, 1, 4, 5, 4, 6],
[1, 2, 5, 5, 5, 6],
[6],
[5, 3, 1, 3, 5, 4, 1, 3, 6],
[5, 1, 3, 2, 3, 6],
[4, 1, 3, 2, 5, 3, 2, 6],
[1, 3, 3, 1, 6],
[4, 3, 3, 4, 5, 3, 3, 2, 5, 2, 1, 1, 4, 4, 6],
[1, 3, 1, 1, 6],
[4, 1, 4, 3, 4, 2, 4, 2, 3, 2, 6],
[5, 4, 4, 4, 3, 2, 5, 1, 6],
[1, 5, 5, 4, 5, 1, 4, 6],
[3, 6],
[2, 2, 2, 5, 3, 2, 3, 1, 1, 3, 4, 2, 1, 6],
[5, 2, 2, 1, 4, 3, 3, 3, 3, 1, 6],
[4, 1, 6],
[1, 4, 5, 5, 4, 2, 3, 2, 6],
[6],
[6],
[5, 3, 4, 2, 1, 6],
[1, 5, 5, 3, 5, 4, 5, 6],
[1, 3, 2, 1, 3, 4, 3, 5, 3, 6],
[5, 2, 1, 5, 5, 2, 3, 4, 5, 5, 3, 2, 2, 6],
[2, 2, 2, 3, 1, 5, 4, 1, 4, 6],
[6],
[6],
[6],
[6],
[1, 1, 3, 4, 5, 1, 3, 6],
[2, 3, 6],
[2, 2, 2, 6],
[4, 2, 5, 4, 2, 2, 4, 3, 3, 4, 5, 1, 6],
[1, 4, 4, 2, 4, 6],
[6],
[6],
[2, 3, 6],
[3, 4, 3, 2, 6],
[4, 6],
[3, 4, 4, 3, 4, 6],
[4, 6],
[2, 4, 2, 4, 5, 5, 1, 5, 2, 5, 6],
[3, 1, 3, 4, 6],
[3, 1, 4, 3, 6],
[3, 3, 5, 2, 2, 3, 1, 3, 6],
[2, 2, 3, 6],
[3,
1,
1,
2,
5,
4,
5,
1,
4,
2,
5,
1,
4,
5,
2,
4,
1,
1,
5,
4,
5,
2,
1,
3,
3,
5,
4,
2,
4,
6],
[3, 1, 4, 6],
[4, 6],
[4, 3, 2, 3, 2, 3, 2, 5, 5, 3, 1, 2, 2, 3, 6],
[6],
[4, 5, 4, 4, 6],
[2, 4, 2, 2, 6],
[4, 6],
[5, 4, 1, 3, 1, 6],
[6],
[5, 2, 2, 1, 3, 1, 3, 1, 2, 4, 3, 1, 1, 6],
[2, 2, 1, 5, 3, 5, 5, 6],
[3,
4,
3,
2,
3,
4,
1,
2,
3,
4,
1,
3,
2,
4,
2,
2,
3,
2,
1,
4,
4,
4,
2,
4,
2,
6],
[5, 2, 2, 4, 3, 3, 3, 4, 3, 6],
[4, 1, 6],
[4, 2, 1, 4, 1, 1, 3, 4, 5, 3, 6],
[6],
[2, 6],
[4, 1, 5, 5, 5, 5, 1, 2, 1, 5, 4, 6],
[1, 3, 6],
[1, 5, 2, 5, 1, 3, 5, 2, 3, 2, 6],
[6],
[2, 1, 1, 1, 4, 5, 5, 3, 3, 6],
[2, 4, 6],
[6],
[4, 5, 4, 4, 5, 1, 5, 6],
[1, 2, 6],
[6],
[3, 3, 6],
[5, 4, 4, 3, 3, 5, 1, 3, 4, 1, 4, 3, 5, 1, 5, 6],
[5, 6],
[1, 2, 6],
[5, 4, 2, 6],
[2, 2, 5, 5, 5, 6],
[5, 3, 1, 6],
[3, 4, 6],
[6],
[1, 4, 2, 4, 1, 4, 2, 1, 4, 6],
[4, 4, 1, 6],
[4, 6],
[5, 6],
[5, 1, 1, 6],
[4, 2, 1, 2, 6],
[3, 5, 2, 5, 5, 1, 2, 2, 2, 2, 5, 5, 4, 3, 1, 3, 1, 6],
[1, 3, 4, 3, 5, 4, 1, 2, 2, 5, 4, 2, 3, 5, 4, 6],
[5, 1, 4, 6],
[5, 6],
[1, 6],
[4, 5, 1, 4, 6],
[1, 1, 3, 4, 5, 5, 3, 5, 1, 4, 6]]
Now we have a bunch of results. Python can count them for us.
In [174]:
counted_throws = []
for trial in results:
throws = len(trial)
counted_throws.append(throws)
counted_throws
Out[174]:
[13,
5,
2,
1,
8,
13,
6,
1,
9,
6,
8,
5,
15,
5,
11,
9,
8,
2,
14,
11,
3,
9,
1,
1,
6,
8,
10,
14,
10,
1,
1,
1,
1,
8,
3,
4,
13,
6,
1,
1,
3,
5,
2,
6,
2,
11,
5,
5,
9,
4,
30,
4,
2,
15,
1,
5,
5,
2,
6,
1,
14,
8,
26,
10,
3,
11,
1,
2,
12,
3,
11,
1,
10,
3,
1,
8,
3,
1,
3,
16,
2,
3,
4,
6,
4,
3,
1,
10,
4,
2,
2,
4,
5,
18,
16,
4,
2,
2,
5,
11]
We can see that on the first trial we had to throw the dice 13 times before getting a 6
In [175]:
len(results[0])
Out[175]:
13
How many times in total did we throw the dice?
In [176]:
sum(counted_throws)
Out[176]:
632
We threw it 632 times! Wow, that’s a lot.
But how many times on average? We’ll we had 100 trials? How much is that?
In [178]:
sum(counted_throws) / float(trials) # See that 'float' just ignore it for the time being.
Out[178]:
6.32
Humm… so, its taken us about 6.3 throws on average to get a 6.
Let’s see what happens if we throw more than 100 times.
But, before we do our experiment, let’s writes some code to simplify our testing.
In [179]:
# This code is similar to the code above, but I changed some variable names to make it easier to read.
def average_throws_till_6(trials):
throws = []
for i in range(trials):
trial = toss_till_6()
throws.append(trial)
counted_throws = []
for throw in throws:
counted_throws.append(len(throw))
average = sum(counted_throws) / float(trials)
return average
In [180]:
trials = 1000
average_throws_till_6(trials)
Out[180]:
5.906
Wow. 5.906 throws on average.
When we did it with 10 trials we got 4.8 throws on average….
100 throws was 6.3 throws on average.
1000 throws got 5.906 throws on average.
But why stop at 1000 throws… let’s go higher! Let’s push the limits. It’s not like our arm will fall off from throwing the dice.
In [181]:
trials = 10000 # That's ten thousand throws!
average_throws_till_6(trials)
Out[181]:
5.9077
Throws on average with 10,000 throws is 5.9077.
It looks like we’re getting pretty close to 6.
What will happen if we do it 50,000 times? What about 100,000 times?
In [182]:
trials = 50000
average_throws_till_6(trials)
Out[182]:
6.02692
In [183]:
trials = 100000
average_throws_till_6(trials)
Out[183]:
5.98154
It looks like as we go a lot higher the number doesn’t change much.
How many tosses, on average, does it take to get a 6?
I guess it takes about 6 throws on average until we reach the number 6
It seems that we’ve answered our question, the answer is 6. Right?
But what if we change the question.
How many throws, on average does it take to get a 3?
Note: I just decided on 3, it could have been a different number. But, maybe we should test other numbers too?
Our original code was written to only test for a 6, so let’s change to to test for any number.
In [184]:
# our original one was called "toss_til_6" we've renamed the 6 to "value"
# Now we can test numbers other than 6, like 3.
def toss_till_value(value):
tosses = []
while True:
face = random.randint(1, 6)
tosses.append(face)
if face == value:
break
return tosses
How about we try out our new code by running an experiment with 3.
In [185]:
toss_till_value(3)
Out[185]:
[5, 4, 2, 1, 4, 3]
We had to toss the dice 6 times until we got a 3.
What happens if we toss it a bunch of times like before?
Let’s create some code to do the calculations.
In [187]:
# This code is similar to the code above, but made a couple changes...
# "toss_till_6" is now "toss_till_value"
# we have to supply a value
def average_throws_till_value(trials, value):
throws = []
for i in range(trials):
trial = toss_till_value(value)
throws.append(trial)
counted_throws = []
for throw in throws:
counted_throws.append(len(throw))
average = sum(counted_throws) / float(trials)
return average
In [188]:
average_throws_till_value(10, 3)
Out[188]:
7.1
Humm, when we do 10 throws it takes 7.1 throws on average until we get a 3.
What happens if we do it more and more!
In [189]:
average_throws_till_value(1000, 3)
Out[189]:
5.916
In [190]:
average_throws_till_value(10000, 3)
Out[190]:
5.9771
In [191]:
average_throws_till_value(10000, 3)
Out[191]:
6.0047
In [192]:
average_throws_till_value(1000000, 3)
Out[192]:
5.996575
These numbers are pretty close to 6, especially when we do a high number of throws.
We’ve done 3 and 6, both take about 6 throws on average. What about the remaining faces…
1. ???
2. ???
4. ???
5. ???
Testing 1, 2, 4, 5
In [194]:
print "1:", average_throws_till_value(10000, 1)
print "2:", average_throws_till_value(10000, 2)
print "3:", average_throws_till_value(10000, 3)
print "4:", average_throws_till_value(10000, 4)
print "5:", average_throws_till_value(10000, 5)
print "6:", average_throws_till_value(10000, 6)
1: 5.9834
2: 6.0456
3: 6.0248
4: 5.9496
5: 5.9938
6: 5.9935
I guess it doesn’t matter what face value we choose. It takes 6 throws on average to get any face!
# What if…
• We tried a dice with 6 sides, but what if we used one with 10 sides? 15 sides? 20 sides? 20-sided dice exist!
• What if we had a die that had 6 sides but instead of 1,2,3,4,5,6 they were 1,2,3,3,3,4? (See, it doesn’t even go to 6!)
• Or a 10 sided die that went to 6 but had sides like 1,1,1,2,2,3,4,5,6,6 ? | 8,115 | 19,366 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2013-48 | longest | en | 0.724225 |
https://samacheerkalvi.guru/samacheer-kalvi-10th-maths-chapter-8-ex-8-2/ | 1,716,446,791,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00071.warc.gz | 434,995,067 | 24,897 | # Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.2
## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.2
Question 1.
The standard deviation and coefficient of variation of a data are 1.2 and 25.6 respectively. Find the value of mean.
Solution:
Co-efficient of variation C.V. = $$\mathrm{C.V}=\frac{\sigma}{\overline{x}} \times 100$$
Question 2.
The standard deviation and coefficient of variation of a data are 1.2 and 25.6 respectively. Find the value of mean.
Solution:
Question 3.
If the mean and coefficient of variation of a data are 15 and 48 respectively, then find the value of standard deviation.
Solution:
Question 4.
If n = 5 , $$\overline{x}$$ = 6, $$\Sigma x^{2}$$ = 765, then calculate the coefficient of variation.
Solution:
Question 5.
Find the coefficient of variation of 24, 26, 33, 37, 29,31.
Solution:
Question 6.
The time taken (in minutes) to complete a homework by 8 students in a day are given by 38, 40, 47, 44, 46, 43, 49, 53. Find the coefficient of variation.
Solution:
Question 7.
The total marks scored by two students Sathya and Vidhya in 5 subjects are 460 and 480 with standard deviation 4.6 and 2.4 respectively. Who is more consistent in performance?
Solution:
Question 8.
The mean and standard deviation of marks obtained by 40 students of a class in three subjects Mathematics, Science and Social Science are given below.
Which of the three subjects shows highest variation and which shows lowest variation in marks?
Solution:
Science subject shows highest variation. Social science shows lowest variation.
Question 9.
The temperature of two cities A and B in a winter season are given below.
Find which city is more consistent in temperature changes?
Solution:
∴ Co-efficient of variation of City A is less than C.V of City B.
∴ City A is more consistent. | 479 | 1,879 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-22 | latest | en | 0.888058 |
https://passemall.com/study-guide-for-tsi-math/ | 1,670,348,944,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711111.35/warc/CC-MAIN-20221206161009-20221206191009-00135.warc.gz | 480,716,285 | 18,792 | # Ultimate Study Guide for TSI Math [2022 Updated]
In order to ace TSI math, having a comprehensive and effective study guide for TSI math is very important. Check this article for more beneficial information!
November 15, 2022
Before diving into the study guide for TSI math, let’s discover what the TSI test is. The Texas Success Initiative Assessment, often known as the TSI, is a test that is given to incoming students in Texas to evaluate how well they are prepared for the amount of work required in college. Every prospective student at the Texas college is required to take the TSI test unless they meet the requirements to get an exemption under certain conditions. The results of this test indicate whether or not a student needs to participate in remedial teaching methods.
On this website, we also offer thousands of free TSI practice test questions for Math and other TSI sections to help you thoroughly prepare for this exam!
The TSI examination is a multiple-choice, computer-adaptive test that does not have a time limit. When the test is concluded, the student will immediately be provided with both TSI scores and specific information about their current level of skill.
## TSI Math Format
The mathematics portion of the TSI Assessment consists of around twenty questions with multiple-choice answers. If a student does not get the minimum score necessary, they will be given an extra 40 problems to do so that their mathematical ability may be evaluated. The following 40 questions will cover more fundamental aspects of mathematics, such as percentages, decimals, area, and perimeter, along with the operation of multiplication.
On the TSI Assessment Math porting, you will only be asked questions with multiple-choice answers. The majority of the material in Algebra 1 and Algebra 2 is covered by these questions. If you want to be successful in those areas, you need to put your attention there.
Geometry, algebraic functions, basic and intermediate algebra, data analysis, statistics, and probability are all subjects that you should be acquainted with. On the TSI Math subtest, a student’s comprehension of measures (such as planar geometry), transformations and symmetry, area, volume, and three-dimensional measurements are also tested.
To get a passing score on the mathematics portion of the TSI Assessment, a student has to have an understanding of mathematical functions, basic operations, rational and exponential expressions, as well as mathematical equations.
In addition, a student’s skills and level of comprehension in the areas of probability, statistics, and data analysis are examined and scored. Participants in the TSI test are required to demonstrate an understanding of probabilistic reasoning, the ability to analyze categorical and quantitative data, as well as statistical data and metrics.
## What’s on TSI Math?
Mathematics, reading comprehension, and writing are going to be tested in the TSI exam. In the arithmetic portion of this test, there will be a total of 20 questions with multiple-choice answers. Within this part, students will be evaluated on the following four key topics:
• Elementary Algebra and Functions
• Intermediate Algebra and Functions
• Geometry and Measurement
• Data Analysis, Statistics, and Probability
### Elementary Algebra & Functions
The skill of test-takers in solving a range of algebraic problems at both the elementary and intermediate levels is evaluated during the algebra section of the examination. In the elementary algebraic expressions and functions portion of the test, participants will be asked to combine or simplify algebraic expressions and solve basic equations. Additionally, they will be tested on their ability to solve simple equations.
### Intermediate Algebra & Functions
Participants in the Intermediate Algebra and Functions test are expected to have an understanding of what a quadratic equation is as well as how to solve a quadratic equation. In this section, it is essential to learn how to simplify square roots and how to use the quadratic formula to discover answers to problems.
Additionally, it is important to know how to simplify square roots. In the section of the test that covers intermediate algebra and functions, there is supplemental material in the form of additional questions on radicals, functions, and equations that are rational and exponential.
### Geometry and Measurement
In the geometry and measurement section, test takers’ understanding of geometric concepts such as area, symmetry, and three-dimensional shape will be tested. Participants in the examination need to be prepared to answer questions such as those on missing dimensions for rectangles, triangles, and circles in this section.
The questions included in this section are designed to evaluate the candidates’ knowledge of solid figures, three-dimensional figures, and geometric modeling for use in computer applications.
### Data Analysis, Statistics, and Probability
The capacity of students to deal with data sets, probabilities, and statistics is tested in the section of the test that focuses on data analysis, statistics, and probability. It is very necessary to have a solid understanding of how to do data analysis on the tables and charts included in this section.
In this part, you will find questions not just on the mean and the median, but also about a variety of other statistical and probabilistic topics.
## What is the TSI Math Passing Score?
To pass the Math TSI, you need a score of 950 or an ABE diagnostic level of 6, whichever is higher. The ELAR subcomponent is more challenging to assess accurately. To pass the ELAR portion, you need to have a placement score of 945 and a score of 5 on the essay. If your total score is lower than 945, you are required to have an ABE diagnostic level of 5, as well as a score of 5 on your essay.
It is very necessary to grasp that the TSI Assessment is not a test with a pass/fail grading system. The term “passing score” refers to the benchmark figure that is used to evaluate pupils’ readiness for college.
## How to Study for the TSI Math Test? Study Guide for TSI Math
### Prepare effectively
The most successful preparation for the TSI math test is one that is both thorough and efficient. Because your time is valuable, we will endeavor to provide explanations that are condensed, clear, and devoid of fluff. If a hypothesis has not been investigated using the TSI, then it should not be included in your research.
Make use of diagnostic pre-tests so that you can figure out what you already know and what you don’t know. Because of this, you will be better able to focus on the TSI preparation material that will have the most significant impact on how well you do on the test.
There is no way around the fact that some of the pupils won’t be able to sail through the class. If you haven’t been to school in a long time, or if you’ve never been good at math and don’t know how to pass the TSI math test, you may need to go back over the basics. This is especially true if you haven’t been successful in the past.
Students who have a significant need for help must have easy access to foundational courses that emphasize core concepts. If you feel as if you need to brush up on your core knowledge before moving on to more complex topics, using this component of the TSI exam preparation might prove to be extremely beneficial. | 1,462 | 7,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-49 | longest | en | 0.938021 |
https://doc.sagemath.org/html/en/reference/combinat/sage/combinat/chas/fsym.html | 1,709,196,260,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474795.48/warc/CC-MAIN-20240229071243-20240229101243-00340.warc.gz | 213,317,862 | 18,649 | # Poirier-Reutenauer Hopf algebra of standard tableaux#
AUTHORS:
• Franco Saliola (2012): initial implementation
• Travis Scrimshaw (2018-04-11): added missing doctests and reorganization
class sage.combinat.chas.fsym.FSymBases(parent_with_realization)#
The category of graded bases of $$FSym$$ and $$FSym^*$$ indexed by standard tableaux.
class ElementMethods#
Bases: object
duality_pairing(other)#
Compute the pairing between self and an element other of the dual.
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: G = FSym.G()
sage: F = G.dual_basis()
sage: elt = G[[1,3],[2]] - 3*G[[1,2],[3]]
sage: elt.duality_pairing(F[[1,3],[2]])
1
sage: elt.duality_pairing(F[[1,2],[3]])
-3
sage: elt.duality_pairing(F[[1,2]])
0
class ParentMethods#
Bases: object
basis(degree=None)#
The basis elements (optionally: of the specified degree).
OUTPUT: Family
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: TG = FSym.G()
sage: TG.basis()
Lazy family (Term map from Standard tableaux to Hopf algebra of standard tableaux
over the Rational Field in the Fundamental basis(i))_{i in Standard tableaux}
sage: TG.basis().keys()
Standard tableaux
sage: TG.basis(degree=3).keys()
Standard tableaux of size 3
sage: TG.basis(degree=3).list()
[G[123], G[13|2], G[12|3], G[1|2|3]]
degree_on_basis(t)#
Return the degree of a standard tableau in the algebra of free symmetric functions.
This is the size of the tableau t.
EXAMPLES:
sage: G = algebras.FSym(QQ).G()
sage: t = StandardTableau([[1,3],[2]])
sage: G.degree_on_basis(t)
3
sage: u = StandardTableau([[1,3,4,5],[2]])
sage: G.degree_on_basis(u)
5
duality_pairing(x, y)#
The canonical pairing between $$FSym$$ and $$FSym^*$$.
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: G = FSym.G()
sage: F = G.dual_basis()
sage: t1 = StandardTableau([[1,3,5],[2,4]])
sage: t2 = StandardTableau([[1,3],[2,5],[4]])
sage: G.duality_pairing(G[t1], F[t2])
0
sage: G.duality_pairing(G[t1], F[t1])
1
sage: G.duality_pairing(G[t2], F[t2])
1
sage: F.duality_pairing(F[t2], G[t2])
1
sage: z = G[[1,3,5],[2,4]]
sage: all(F.duality_pairing(F[p1] * F[p2], z) == c
....: for ((p1, p2), c) in z.coproduct())
True
duality_pairing_matrix(basis, degree)#
The matrix of scalar products between elements of $$FSym$$ and elements of $$FSym^*$$.
INPUT:
• basis – a basis of the dual Hopf algebra
• degree – a non-negative integer
OUTPUT:
• the matrix of scalar products between the basis self and the basis basis in the dual Hopf algebra of degree degree
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: G = FSym.G()
sage: G.duality_pairing_matrix(G.dual_basis(), 3)
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[0 0 0 1]
one_basis()#
Return the basis index corresponding to $$1$$.
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: TG = FSym.G()
sage: TG.one_basis()
[]
super_categories()#
The super categories of self.
EXAMPLES:
sage: from sage.combinat.chas.fsym import FSymBases
sage: FSym = algebras.FSym(ZZ)
sage: bases = FSymBases(FSym)
sage: bases.super_categories()
[Category of realizations of Hopf algebra of standard tableaux over the Integer Ring,
Join of Category of realizations of hopf algebras over Integer Ring
and Category of graded algebras over Integer Ring
and Category of graded coalgebras over Integer Ring,
Category of graded connected hopf algebras with basis over Integer Ring]
Abstract base class for graded bases of $$FSym$$ and of $$FSym^*$$ indexed by standard tableaux.
This must define the following attributes:
• _prefix – the basis prefix
some_elements()#
Return some elements of self.
EXAMPLES:
sage: G = algebras.FSym(QQ).G()
sage: G.some_elements()
[G[], G[1], G[12], G[1] + G[1|2], G[] + 1/2*G[1]]
class sage.combinat.chas.fsym.FreeSymmetricFunctions(base_ring)#
The free symmetric functions.
The free symmetric functions is a combinatorial Hopf algebra defined using tableaux and denoted $$FSym$$.
Consider the Hopf algebra $$FQSym$$ (FreeQuasisymmetricFunctions) over a commutative ring $$R$$, and its bases $$(F_w)$$ and $$(G_w)$$ (where $$w$$, in both cases, ranges over all permutations in all symmetric groups $$S_0, S_1, S_2, \ldots$$). For each word $$w$$, let $$P(w)$$ be the P-tableau of $$w$$ (that is, the first of the two tableaux obtained by applying the RSK algorithm to $$w$$; see RSK()). If $$t$$ is a standard tableau of size $$n$$, then we define $$\mathcal{G}_t \in FQSym$$ to be the sum of the $$F_w$$ with $$w$$ ranging over all permutations of $$\{1, 2, \ldots, n\}$$ satisfying $$P(w) = t$$. Equivalently, $$\mathcal{G}_t$$ is the sum of the $$G_w$$ with $$w$$ ranging over all permutations of $$\{1, 2, \ldots, n\}$$ satisfying $$Q(w) = t$$ (where $$Q(w)$$ denotes the Q-tableau of $$w$$).
The $$R$$-linear span of the $$\mathcal{G}_t$$ (for $$t$$ ranging over all standard tableaux) is a Hopf subalgebra of $$FQSym$$, denoted by $$FSym$$ and known as the free symmetric functions or the Poirier-Reutenauer Hopf algebra of tableaux. It has been introduced in [PoiReu95], where it was denoted by $$(\ZZ T, \ast, \delta)$$. (What we call $$\mathcal{G}_t$$ has just been called $$t$$ in [PoiReu95].) The family $$(\mathcal{G}_t)$$ (with $$t$$ ranging over all standard tableaux) is a basis of $$FSym$$, called the Fundamental basis.
EXAMPLES:
As explained above, $$FSym$$ is constructed as a Hopf subalgebra of $$FQSym$$:
sage: G = algebras.FSym(QQ).G()
sage: F = algebras.FQSym(QQ).F()
sage: G[[1,3],[2]]
G[13|2]
sage: G[[1,3],[2]].to_fqsym()
G[2, 1, 3] + G[3, 1, 2]
sage: F(G[[1,3],[2]])
F[2, 1, 3] + F[2, 3, 1]
This embedding is a Hopf algebra morphism:
sage: all(F(G[t1] * G[t2]) == F(G[t1]) * F(G[t2])
....: for t1 in StandardTableaux(2)
....: for t2 in StandardTableaux(3))
True
sage: FF = F.tensor_square()
sage: all(FF(G[t].coproduct()) == F(G[t]).coproduct()
....: for t in StandardTableaux(4))
True
There is a Hopf algebra map from $$FSym$$ onto the Hopf algebra of symmetric functions, which maps a tableau $$t$$ to the Schur function indexed by the shape of $$t$$:
sage: TG = algebras.FSym(QQ).G()
sage: t = StandardTableau([[1,3],[2,4],[5]])
sage: TG[t]
G[13|24|5]
sage: TG[t].to_symmetric_function()
s[2, 2, 1]
The Hopf algebra of tableaux on the Fundamental basis.
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: TG = FSym.G()
sage: TG
Hopf algebra of standard tableaux over the Rational Field
in the Fundamental basis
Elements of the algebra look like:
sage: TG.an_element()
2*G[] + 2*G[1] + 3*G[12]
class Element#
to_fqsym()#
Return the image of self under the natural inclusion map to $$FQSym$$.
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: G = FSym.G()
sage: t = StandardTableau([[1,3],[2,4],[5]])
sage: G[t].to_fqsym()
G[2, 1, 5, 4, 3] + G[3, 1, 5, 4, 2] + G[3, 2, 5, 4, 1]
+ G[4, 1, 5, 3, 2] + G[4, 2, 5, 3, 1]
to_symmetric_function()#
Return the image of self under the natural projection map to $$Sym$$.
The natural projection map $$FSym \to Sym$$ sends each standard tableau $$t$$ to the Schur function $$s_\lambda$$, where $$\lambda$$ is the shape of $$t$$. This map is a surjective Hopf algebra homomorphism.
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: G = FSym.G()
sage: t = StandardTableau([[1,3],[2,4],[5]])
sage: G[t].to_symmetric_function()
s[2, 2, 1]
coproduct_on_basis(t)#
Return the coproduct of the basis element indexed by t.
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: G = FSym.G()
sage: t = StandardTableau([[1,2,5], [3,4]])
sage: G.coproduct_on_basis(t)
G[] # G[125|34] + G[1] # G[12|34] + G[1] # G[124|3]
+ G[1|2] # G[13|2] + G[12] # G[12|3] + G[12] # G[123]
+ G[12|34] # G[1] + G[123] # G[12] + G[125|34] # G[]
+ G[13|2] # G[1|2] + G[13|2] # G[12] + G[134|2] # G[1]
dual_basis()#
Return the dual basis to self.
EXAMPLES:
sage: G = algebras.FSym(QQ).G()
sage: G.dual_basis()
Dual Hopf algebra of standard tableaux over the Rational Field
in the FundamentalDual basis
product_on_basis(t1, t2)#
Return the product of basis elements indexed by t1 and t2.
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: G = FSym.G()
sage: t1 = StandardTableau([[1,2], [3]])
sage: t2 = StandardTableau([[1,2,3]])
sage: G.product_on_basis(t1, t2)
G[12456|3] + G[1256|3|4] + G[1256|34] + G[126|35|4]
sage: t1 = StandardTableau([[1],[2]])
sage: t2 = StandardTableau([[1,2]])
sage: G.product_on_basis(t1, t2)
G[134|2] + G[14|2|3]
sage: t1 = StandardTableau([[1,2],[3]])
sage: t2 = StandardTableau([[1],[2]])
sage: G.product_on_basis(t1, t2)
G[12|3|4|5] + G[12|34|5] + G[124|3|5] + G[124|35]
G#
alias of Fundamental
a_realization()#
Return a particular realization of self (the Fundamental basis).
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: FSym.a_realization()
Hopf algebra of standard tableaux over the Rational Field
in the Fundamental basis
dual()#
Return the dual Hopf algebra of $$FSym$$.
EXAMPLES:
sage: algebras.FSym(QQ).dual()
Dual Hopf algebra of standard tableaux over the Rational Field
class sage.combinat.chas.fsym.FreeSymmetricFunctions_Dual(base_ring)#
The Hopf dual $$FSym^*$$ of the free symmetric functions $$FSym$$.
See FreeSymmetricFunctions for the definition of the latter.
Recall that the fundamental basis of $$FSym$$ consists of the elements $$\mathcal{G}_t$$ for $$t$$ ranging over all standard tableaux. The dual basis of this is called the dual fundamental basis of $$FSym^*$$, and is denoted by $$(\mathcal{G}_t^*)$$. The Hopf dual $$FSym^*$$ is isomorphic to the Hopf algebra $$(\ZZ T, \ast', \delta')$$ from [PoiReu95]; the isomorphism sends a basis element $$\mathcal{G}_t^*$$ to $$t$$.
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: TF = FSym.dual().F()
sage: TF[1,2] * TF[[1],[2]]
F[12|3|4] + F[123|4] + F[124|3] + F[13|2|4] + F[134|2] + F[14|2|3]
sage: TF[[1,2],[3]].coproduct()
F[] # F[12|3] + F[1] # F[1|2] + F[12] # F[1] + F[12|3] # F[]
The Hopf algebra $$FSym^*$$ is a Hopf quotient of $$FQSym$$; the canonical projection sends $$F_w$$ (for a permutation $$w$$) to $$\mathcal{G}_{Q(w)}^*$$, where $$Q(w)$$ is the Q-tableau of $$w$$. This projection is implemented as a coercion:
sage: FQSym = algebras.FQSym(QQ)
sage: F = FQSym.F()
sage: TF(F[[1, 3, 2]])
F[12|3]
sage: TF(F[[5, 1, 4, 2, 3]])
F[135|2|4]
F#
alias of FundamentalDual
The dual to the Hopf algebra of tableaux, on the fundamental dual basis.
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: TF = FSym.dual().F()
sage: TF
Dual Hopf algebra of standard tableaux over the Rational Field
in the FundamentalDual basis
Elements of the algebra look like:
sage: TF.an_element()
2*F[] + 2*F[1] + 3*F[12]
class Element#
to_quasisymmetric_function()#
Return the image of self under the canonical projection $$FSym^* \to QSym$$ to the ring of quasi-symmetric functions.
This projection is the adjoint of the canonical injection $$NSym \to FSym$$ (see to_fsym()). It sends each tableau $$t$$ to the fundamental quasi-symmetric function $$F_\alpha$$, where $$\alpha$$ is the descent composition of $$t$$.
EXAMPLES:
sage: F = algebras.FSym(QQ).dual().F()
sage: F[[1,3,5],[2,4]].to_quasisymmetric_function()
F[1, 2, 2]
coproduct_on_basis(t)#
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: TF = FSym.dual().F()
sage: t = StandardTableau([[1,2,5], [3,4]])
sage: TF.coproduct_on_basis(t)
F[] # F[125|34] + F[1] # F[134|2] + F[12] # F[123]
+ F[12|3] # F[12] + F[12|34] # F[1] + F[125|34] # F[]
dual_basis()#
Return the dual basis to self.
EXAMPLES:
sage: F = algebras.FSym(QQ).dual().F()
sage: F.dual_basis()
Hopf algebra of standard tableaux over the Rational Field
in the Fundamental basis
product_on_basis(t1, t2)#
EXAMPLES:
sage: FSym = algebras.FSym(QQ)
sage: TF = FSym.dual().F()
sage: t1 = StandardTableau([[1,2]])
sage: TF.product_on_basis(t1, t1)
F[12|34] + F[123|4] + F[1234] + F[124|3] + F[13|24] + F[134|2]
sage: t0 = StandardTableau([])
sage: TF.product_on_basis(t1, t0) == TF[t1] == TF.product_on_basis(t0, t1)
True
a_realization()#
Return a particular realization of self (the Fundamental dual basis).
EXAMPLES:
sage: FSym = algebras.FSym(QQ).dual()
sage: FSym.a_realization()
Dual Hopf algebra of standard tableaux over the Rational Field
in the FundamentalDual basis
dual()#
Return the dual Hopf algebra of self, which is $$FSym$$.
EXAMPLES:
sage: D = algebras.FSym(QQ).dual()
sage: D.dual()
Hopf algebra of standard tableaux over the Rational Field
sage.combinat.chas.fsym.ascent_set(t)#
Return the ascent set of a standard tableau t (encoded as a sorted list).
The ascent set of a standard tableau $$t$$ is defined as the set of all entries $$i$$ of $$t$$ such that the number $$i+1$$ either appears to the right of $$i$$ or appears in a row above $$i$$ or does not appear in $$t$$ at all.
EXAMPLES:
sage: from sage.combinat.chas.fsym import ascent_set
sage: t = StandardTableau([[1,3,4,7],[2,5,6],[8]])
sage: ascent_set(t)
[2, 3, 5, 6, 8]
sage: ascent_set(StandardTableau([]))
[]
sage: ascent_set(StandardTableau([[1, 2, 3]]))
[1, 2, 3]
sage: ascent_set(StandardTableau([[1, 2, 4], [3]]))
[1, 3, 4]
sage: ascent_set([[1, 3, 5], [2, 4]])
[2, 4, 5]
sage.combinat.chas.fsym.descent_composition(t)#
Return the descent composition of a standard tableau t.
This is the composition of the size of $$t$$ whose partial sums are the elements of the descent set of t (see descent_set()).
EXAMPLES:
sage: from sage.combinat.chas.fsym import descent_composition
sage: t = StandardTableau([[1,3,4,7],[2,5,6],[8]])
sage: descent_composition(t)
[1, 3, 3, 1]
sage: descent_composition([[1, 3, 5], [2, 4]])
[1, 2, 2]
sage.combinat.chas.fsym.descent_set(t)#
Return the descent set of a standard tableau t (encoded as a sorted list).
The descent set of a standard tableau $$t$$ is defined as the set of all entries $$i$$ of $$t$$ such that the number $$i+1$$ appears in a row below $$i$$ in $$t$$.
EXAMPLES:
sage: from sage.combinat.chas.fsym import descent_set
sage: t = StandardTableau([[1,3,4,7],[2,5,6],[8]])
sage: descent_set(t)
[1, 4, 7]
sage: descent_set(StandardTableau([]))
[]
sage: descent_set(StandardTableau([[1, 2, 3]]))
[]
sage: descent_set(StandardTableau([[1, 2, 4], [3]]))
[2]
sage: descent_set([[1, 3, 5], [2, 4]])
[1, 3]
sage.combinat.chas.fsym.standardize(t)#
Return the standard tableau corresponding to a given semistandard tableau t with no repeated entries.
Note
This is an optimized version of Tableau.standardization() for computations in $$FSym$$ by using the assumption of no repeated entries in t.
EXAMPLES:
sage: from sage.combinat.chas.fsym import standardize
sage: t = Tableau([[1,3,5,7],[2,4,8],[9]])
sage: standardize(t)
[[1, 3, 5, 6], [2, 4, 7], [8]]
sage: t = Tableau([[3,8,9,15],[5,10,12],[133]])
sage: standardize(t)
[[1, 3, 4, 7], [2, 5, 6], [8]] | 4,995 | 14,607 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | longest | en | 0.483409 |
https://www.renskevents.nl/14/05/13/1177/weight-of-river-sand.html | 1,555,977,715,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578583000.29/warc/CC-MAIN-20190422235159-20190423021159-00507.warc.gz | 803,414,172 | 7,585 | weight of river sand
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The primary use for river sand is for pre-loading because of its relatively light weight per cubic metre. When compared to other sand products, like bank sand or pit sand, river sand holds a 10-20 per cent weight advantage. | 2,257 | 9,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-18 | longest | en | 0.887638 |
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Say I have a higher-order-function, ‘applier’, that takes another function and some arguments, applies the function to the arguments and returns the results multiplied by two:
`Func<Func<int, int, int>, int, int, int> applier = (f, x, y) => f(x, y) * 2;`
Now let’s say we have a function, ‘multiplier’ that simply multiplies two numbers:
`Func<int, int, int> multiplier = (a, b) => a*b;`
How can we combine applier and multiplier so that we end up with a function that simply multiplies two numbers and doubles the result?
First we curry applier:
`var curried = Functional.Curry(applier);`
So now ‘curried’ looks something like this:
`Func<Func<int, int, int>, Func<int, Func<int, int>>> curried = f => x => y => f(x, y)*2;`
It’s a function that takes a function and returns a function. So if we call it passing in ‘multiplier’ we should get back a function that does what we want:
`var curriedCombined = curried(multiplier);`
The only problem with ‘curriedCombined’ is that it’s in this curried form:
`Func<int, Func<int, int>> curriedCombined = x => y => (x*y)*2;`
So let’s DeCurry it (is there a better term for this?):
`var combined = Functional.DeCurry(curriedCombined);`
Now we have our target function. We can call it just like a normal delegate with two arguments:
`Console.WriteLine("combined(4) = {0}", combined(4, 4));`
Which outputs 32 as expected.
So now we have a nice mechanical way of turning higher-order-functions into normal functions by doing the following:
1. Curry the higher-order-function
2. Pass the function(s) that supply the function argument(s) into the curried version.
3. DeCurry
All I have to do now is some serious reflection and I’ll have a working CurryFacility.
Here is my Functional class (stolen from Oliver Sturm)
`using System;namespace Mike.AdvancedWindsorTricks.Model{ public class Functional { public static Func<T1, Func<T2, T3>> Curry<T1, T2, T3>(Func<T1, T2, T3> function) { return a => b => function(a, b); } public static Func<T1, Func<T2, Func<T3, T4>>> Curry<T1, T2, T3, T4>(Func<T1, T2, T3, T4> function) { return a => b => c => function(a, b, c); } public static Func<T1, Func<T2, Func<T3, Func<T4, T5>>>> Curry<T1, T2, T3, T4, T5>(Func<T1, T2, T3, T4, T5> function) { return a => b => c => d => function(a, b, c, d); } public static Func<T1, T2, T3> DeCurry<T1, T2, T3>(Func<T1, Func<T2, T3>> function) { return (a, b) => function(a)(b); } public static Func<T1, T2, T3, T4> DeCurry<T1, T2, T3, T4>(Func<T1, Func<T2, Func<T3, T4>>> function) { return (a, b, c) => function(a)(b)(c); } public static Func<T1, T2, T3, T4, T5> DeCurry<T1, T2, T3, T4, T5>(Func<T1, Func<T2, Func<T3, Func<T4, T5>>>> function) { return (a, b, c, d) => function(a)(b)(c)(d); } }}`
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Opinions expressed by DZone contributors are their own. | 992 | 3,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-30 | longest | en | 0.830231 |
https://stump.marypat.org/article/1540/mortality-basics-with-meep-median-mode-and-mean-age-at-death-and-life-expectancy | 1,638,339,974,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359093.97/warc/CC-MAIN-20211201052655-20211201082655-00350.warc.gz | 592,128,556 | 5,766 | # STUMP » Articles » Mortality Basics with Meep: Median, Mode, and Mean Age at Death and Life Expectancy » 7 July 2021, 16:26
Where Stu & MP spout off about everything.
### Mortality Basics with Meep: Median, Mode, and Mean Age at Death and Life Expectancy
by
7 July 2021, 16:26
I am writing this in response to something incredibly stupid.
What follows will not be about COVID deaths at all, but just about comparing median age at death, modal age at death (that is, the age when most people die in a given year), mean age at death (aka, average age of everybody who died), and life expectancy.
All four numbers are usually different, and some are very different.
## First important note: calendar year not cohort
When we make these comparisons, at least in the media, it’s based on calendar year experience.
It has nothing to do with cohort, that is following the people born in a specific year. If you want to do retirement planning for yourself, you want to consider cohort, not calendar, mortality for projections.
So for all of the following, I will be using calendar year 2019 experience for the U.S.
I will be looking at whole-number age (no rounding, so Age Last Birthday) for all the stats, because it makes it really easy to do the calculations. I could do something more complicated, like assuming uniform distribution of deaths in a year, but for this demonstration, it’s not as important.
## Age at death distribution for 2019
I pulled this information from WONDER at CDC, which has a full year of 2019 data.
Of the 2,854,838 deaths in 2019, we don’t have the ages of 147 of them. I will just ignore those 147 deaths.
There are also 30,731 deaths at age 100+, and I will treat those as having died at age 100. You will soon see that it won’t make a big difference.
Here is a graph of the deaths for 2019, with key ages labeled. I will explain what they mean below.
## Modal age at death: 87 years old
This one is the easiest to determine just looking at the graph: at what age is the largest number of deaths?
Age 87 just barely edges out age 88 for the mode. There were 76,921 deaths at age 87 in 2019, and 76,891 deaths at age 88.
Obviously, as people get older, the mortality rate gets higher. But, as people die off, there are fewer and fewer people at that age to be hit with that high mortality rate. After the mode, the number of deaths falls off precipitously.
## Median age at death: 77 years old
This is where 50% of the deaths are higher and lower than this age.
This measure is pretty stable (as we shall soon see), in that an uptick in infant mortality or very old age mortality will not have a large effect on this measure. However, it really isn’t very meaningful… nor is the modal age at death. Same for the next measure.
## Mean age at death: 73 years old
The mean age of death is the average age of death over all people who died in 2019. This measure can get pushed or pulled around by early deaths or late deaths. It might be a little higher if I had the full distribution of deaths over age 100, but it wouldn’t move all that much.
But like median and modal ages for death in 2019, the mean age of death tells us little about mortality trends.
The reason that these three measures aren’t very meaningful for considering mortality trends is that they’re all dependent on the number of deaths in a particular age bin in that year.
The number of deaths in each age bin is dependent on the mortality rate for that age and the number of people alive in that bin to begin with. If you have a lot of people in certain age bins compared to others — say a baby boom followed by a baby bust and then a boom again — that will have an effect on the death distribution, even if the mortality rates don’t change. And of course, the mortality rates have changed over time.
## The effect of birth patterns: death spikes
There are two “weird” spikes in the data — the first at age 72, where the number of deaths drops by a lot for age 73 in comparison. What’s up with that?
It’s not because those age 73 have a lower mortality rate than those age 72 — the mortality rate at age 73 is 15% than that at age 72.
But there were 27% fewer people age 73 than age 72 in 2019. Let’s figure out when they were born…. 1946 and 1947.
Going to the CDC:
Notice the two spikes — births dropped at the very beginning of WWII, and then there was a big one-year boom in 1947 (not 1946, actually). The reasoning is very simple: the men went away. The women didn’t have babies. When the men came back… there was some pent-up demand, which spiked in 1947 and then fell back.
By the way, note that fertility rates didn’t peak until the late 1950s.
Immigration didn’t have much of an effect on the age structure of the population, as other nations were similarly affected.
## Life expectancy: 78.8 years old
From the CDC press release in December 2020:
As a result, life expectancy at birth for the U.S. population increased 0.1 year from 2018 to 78.8 years in 2019.
As I’ve mentioned many times before, there is some meaning to tracking the period life expectancy from birth, even if it has little meaning to the person planning their retirement.
Nerdy explanation: Mortality with Meep: Cohort vs. Period Mortality Tables
Note that the life expectancy calculation is not dependent on the age structure of the population at all. It’s just an expected value using the mortality rate from each age bin from that year. Age distribution has no effect on this measure.
Similarly, age-adjusted mortality rates are not affected by the age distribution, because one forces the age distribution to be a certain way to do your weighted average.
## Comparison against 1999
I thought you’d find it interesting to see what the death distribution looked like 20 years before:
mode: 84
median: 77
mean: 72
life expectancy: 76.7
(and those two weird spikes are 20 years earlier)
Note: 1999 median & 2019 median are the same. As I said, the median may not move around a lot, though many things about the death distribution are changing.
Mean moved up 1 year, mode moved up 3 years, and life expectancy moved up 2 years.
## So what does this mean for COVID deaths?
Well, what did the median death age and life expectancy differences tell us in 2019?
Not much, not really. The median depends on the age distribution of the living population in addition to mortality rates. Life expectancy is driven by mortality rates by age alone.
But seriously, comparing life expectancy against the mean, median, or mode tells you very little.
I’m also very suspicious that the stats quoted on Tucker Carlson’s show are right at all (yes, life expectancy dropped, but I don’t think life expectancy in Ohio is that low).
Yes, COVID primarily killed old folks. There was excess mortality for younger folks that was driven by non-COVID causes (which I’ll be getting back to soon — stuff like car accidents and drug overdoses).
I can make that argument based directly on increases in deaths for those age ranges, and comparing against official COVID deaths.
I don’t need to use two entirely different and non-comparable death statistics.
I know Tucker is gonna Tucker, but this is really stupid.
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Mortality Nuggets: NYT Misleads, COVID Deaths Down, and Car Crash Fatalities Up | 1,718 | 7,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-49 | longest | en | 0.956123 |
https://edward-huang.com/functional-programming/2019/12/30/what-is-an-adt-algebraic-data-types/ | 1,606,156,848,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141164142.1/warc/CC-MAIN-20201123182720-20201123212720-00682.warc.gz | 283,087,976 | 7,048 | I got a story to research on whether Circe can polymorphically parse Json objects to ADT. One of the questions that I got was, what is an ADT?
I murmured for a couple of minutes - realizing that I could not say a concrete explanation on what is ADT.
After being called out in front of the standups, I decided that I have to understand what Algebraic Data Types is - and I think this knowledge is essential and is often used in a lot of Functional Programming applications.
To Quote Alvin Alexander -
Algebraic data types are not a way of proactively designing FP (Functional Programming) code; instead, they are a way of categorizing data models you’ve already written, especially FP data models.
We usually know ADT in a lot of FP data models like Enumeration (Enum) type in OOP. However, ADT represents a broader category than just Enums.
1. is that you can do pattern matching style of programming.
2. Your code becomes more “algebra”
## What does “algebra” means?
A lot of us know algebra that we encounter in high school. Example:
1+1 = 2
Those are numeric algebra.
However, as you study more about FP, you get to understand that algebra consists of 2 characteristics:
1. A set of objects
2. Operations that applied to those objects to create new objects
In the numeric algebra case, for instance, 1 are a set of objects of numerical types, and the operators are +. Therefore, 2 is consists of 1 numerical types that applied with an operators +.
There are various “algebra” besides, numerical algebra. There is relational algebra, which is used in the database. In relational algebra, the table represents the set of objects, and operators such as SELECT, AS are represented as the operators. You can see more types of algebra here.
### How is algebra works in programming?
case class Coordinates(x:Int, y:Int)
Chances are, you have already work on ADT even if you don’t know it. The example above represents algebra in programming, because of a set of objects (integers of x, and y), and the constructor’s operators (case class) that creates a new type (Coordinates)
## There are 3 types of ADT in scala:
### Sum Types - usually with sealed trait and a case objects.
Sum types also called enumeration type because you enumerate all the possible instances from the base type. Usually, you phrase those types with “is a” and “or”
### Product Types - case class
Product types are a data type that you create with case class, that has a parameter. Instead of creating a data type whose number of possibilities of concrete instance that you list that extends the base type, it determines the number of possibilities that you have in your constructor fields. Usually, you phrase said those types with a “have a” and “and”
Example:
case class Coordinates(x:Int, y:Int)
How many combinations that need to account for this one?
In this case, it is 2 * 2^256 because there are that many possibilities in an integer.
### Hybrid - both sum and product type, a sealed trait with a case class
Hybrid types follow a sealed trait and a case class.
sealed trait PaymentInstrument
case class Paypal(accountId:String, amount:Int)
case class Ideal(accountId:String, amount:Int)
In this case, Paypal is-a type of PaymentInstrument, or you can phrase it as PaymentInstrument is a Paypal or Ideal.
Then Paypal is a product type where Paypal has-a accountId and amount.
As you noticed, if a case class except a String, there are an infinite amount of possibilities. If a class has an Int, you need to account for 256, and if it accepts a boolean, you need to account for 2 possibilities. Therefore, the fewer possibilities you have to deal with, the amount o code you have to write. One scenario that you should think about when designing your data types.
## 3 Main Takeaways
• Algebraic Data Types is not a way to design your functional programming code, but to categories your data model.
• There are 3 cases of Algebraic Data Types (Sum, Product, and Hybrid)
• After understanding ADT, you also get to realized - more parameters and combination you create in your data types equal more cases that you need to cover. If you need to cover more cases in your code, the unit test is much harder.
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# Common Core Curriculum Framework Training 3 - 5
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## Malissa Jacks
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#### Transcript of Common Core Curriculum Framework Training 3 - 5
"And once I had a teacher who understood. He brought with him the beauty of mathematics. He made me create it for myself. He gave me nothing, and it was more than any other teacher has ever dared to give me." -Cochran
SEDOL Common Core Curriculum Framework Training
Number Sense
Instructional Strategies
Line Models
a/b
1/b
Limit denominators to 2, 3, 4, 6, and 8 (3rd grade)
Limit denominators to 2, 3, 4, 6, 8, 10, 12, and 100 (4th grade)
Area Models
Set Models
Order and Compare
Fractions and Decimals
p. 24
Models - not algorithm
Algorithm
more experience
with models
"benchmark fractions"
3/4 < 5/4
3NF.c includes representation of whole
numbers as fractions (1/1, 2/1, etc.)
Region/Area Models
Line Models
.13 < .2
.134 > .12
.301, .031, .003, .31
Base TEN Fractions
3/10 or 30/100
as .3 or .30
27/10?
ACTIVITY
Place Value
again... always
use visual models
misconception usually occurs with rounding down
400 + 20 + 7 = 427
3 x 100 + 4 x 10 + 7 x 1 + 3 x 1/10 +
9 x 1/100 + 2 x 1/100 = 347.392
Expanded Form
Factors and Multiples
(P. 22)
Strong Connection to....
Multiplication and division
Area
Fractions
Models SHOULD
be used to help students find
factors and multiples
Which numbers had the least amount of arrays?
What numbers have a factor of 2?
What numbers have factors that form a square?
What can you say about the factors for even numbers?
Do even numbers always have two even factors?
Multiplication and Division
(P. 34)
DO NOT...
Have students memorize facts (using drilling)
Teach multiplication and division separately
Teach in chunks of facts (e.g. 5s)
There are three levels of multiplication and division understanding, ALL build off of students' understanding of arrays and repeated addition developed in 2nd grade.
Level 2: Repeated counting on by a given number
Level 3: applying the associative and distributive property to compose and decompose (either into addition/subtraction or multiplication)
All of students’ understanding of multiplication and division situations, levels of representing and solving, and patterns need to culminate by the end of grade 3 in fluent multiplying and dividing.
Level 1: making and counting all quantities involved
Multiplication and Division
(P. 34)
K-2 standards
k-2
Strategies
Making a ten
Decomposing to a ten
Counting on
Related Facts
Using the relationship between addition and subtraction
Properties of Operations
Fluency is NOT automaticity
It does not come from memorization
It is the process of using strategies to solve problems more efficiently
This is something that cannot be assessed through a paper and pencil assessment. Needs to be assessed through an interview based assessment.
Strictly about understanding what multiplication and division mean and how the operations can be applied to real life situations.
Fluency in multiplication and division is much harder than addition and subtraction because there are no general strategies, just patterns and strategies dependent on specific numbers.
Example, when multiplying by 5, the product will always end is a 5 or 0.
(essentially students at this stage are just laying the foundations for standards 2.OA.3 & 2.OA.4)
This is easier for division than multiplication
Easier for multiplication than division
Example:
7 x 6 =
(6 + 1) x 6 =
6 x 6 + 1 x 6 =
36 + 6 = 42
Meaning, at the end of grade 3 all students should be at level 3 or beyond (have facts memorized).
Other key points about teaching multiplication and division.
Multiplication and division must be taught at the same time, NOT as separate units.
Models (especially area/array) are EXTREMELY important in developing understanding of multiplication and division, especially for future fractional concepts.
Present multiplication and division problems in real-life contexts
3 x 50 =
3 x 5 tens =
15 tens =
150
How do they know 15 tens is 150?
Skip counting by 50. 5 tens is 50, so 50, 100, 150.
Counting on by 5 tens. 5 tens is 50, 5 more tens is 100, 5 more tens is 150
Decomposing 15 tens. 15 tens is 10 tens and 5 tens. 10 tens is 100. 5 tens is 50. So, 15 tens is 100 and 50, or 150.
(relies on a strong place value understanding)
How do you move onto long division?
Think of how multi-digit multiplication was introduced.
If a student knows 56 / 8 then they can solve 560 / 8 and 5600 / 8
Our ultimate goal is fluent division...
Place value
Properties of operations
Relationship between multiplication and division
Round UP!
Rounding
Decimal Placement in Multiplication
DO NOT teach the rule of counting digits after a decimal to teach decimal placement.
If students understand place value well, they can reason about the placement of the decimal.
What is a tenth x a tenth?
Students can convert decimals to base-ten fractions
Estimation
Decimal Placement in Division
Connect to multiplication
Use place value reasoning
DO NOT teach students how to count digits after the decimal
Estimation
In mathematics, an algorithm is defined
by its steps and not by the way those
steps are recorded in writing. With this in
mind, minor variations in methods of recording
standard algorithms are acceptable.
remember the pizza's
this can't equal 3/4
3/3 = 1/3 + 1/3 + 1/3
3/3 = 2/3 + 1/3
Like Denominators including Mixed Numbers
this whole idea is built off of always representing
fractions >1 (5/3, 9/4,)
building off that strong k-2 number sense
17/6 - 5/6 or 7/5 +4/5
80% of students on NAEP:
4 1/5 as 21/5, but didn't know 4+1/5
the concept of a mixed number is defined only after
How do we develop students thinking vs. directly teaching
the algorithm?
Add and Subtract with unlike denominators including mixed numbers
Equivalency is the key prerequisite
Estimation should occur before computation
Informal methods should be supported
1/4 + 1/2
3/4 + 2/3
obviously visual fraction models ALWAYS are employed at
the beginning of any new concept so that students construct
their own meaning so that their invented approaches contribute
to the develpment of the algorithm
Nothing states they have to have LCM - they will discover it's easier
How is 5.NF.4 similar to the 4.NF.4a &4.NF.4b standards?
Lets try some problems
without
1 1
_ _
2 3
X
2 4
_ -
3 5
x
Use the manipulatives in front of you to solve the following problem...
Jim is making four pizzas to be shared equally among 5 people. How much pizza does each person get?
This is an example of an equal sharing problem.
How can we use a number line to solve the following problem?
Five people need to run a relay race of four miles, how far does each person have to run if they all run the same distance?
Hunter has 3/4 of a cake left to share among 4 visitors at his house. How much of the leftover cake does each person get?
Cynthia would like to bake cookies. She has one cup of sugar in the pantry. How many batches of cookies can she make using the one cup if each batch requires
A) 1/2 cup B) 1/3 cup C) 2/3 cup D) 3/4 cup
Grade 1 - hour and half hour
- analog and digital
- a.m. and p.m.
Now - nearest minute, solving time intervals (elapsed time)
11:30 12:30 1:30 1:45
The game begins at 11:30 a.m. If it lasts 2 hours and 15 minutes, when will it be over?
8:45 - 11:15
Algebraic Thinking
(P.40)
Do NOT teach students tricks to decode word problems!
This can lead to misconceptions about what operations to use to solve a problem.
This does not create true problem-solving sense.
For example,
Julie has 5 more apples than Lucy. Lucy has 4 apples. How many apples does Julie have?
Lucy has 4 fewer apples than Julie. Lucy has 5 apples. How many apples does Julie have?
What is an arithmetic pattern?
Patterns that change by the same rate, such as adding the same number.
For example, the series 2, 4, 6, 8, 10 is an arithmetic pattern that increases by 2 between each term.
What patterns do you see?
What patterns do you see?
Any sum of two even numbers is even.
Any sum of an even number and an odd number is odd.
The multiples of 4, 6, 8, and 10 are all even because they can all be decomposed into two equal groups.
The doubles (2 addends the same) in an addition table fall on a diagonal while the doubles (multiples of 2) in a multiplication table fall on horizontal and vertical lines.
The multiples of any number fall on a horizontal and a vertical line due to the commutative property.
All the multiples of 5 end in a 0 or 5 while all the multiples of 10 end with 0. Every other multiple of 5 is a multiple of 10.
Any sum of two odd numbers is even.
Why, on a multiplication chart, does the products in each row and column increase by the same amount?
There are 4 beans in a jar. Each day 3 beans are added. How many beans are in the jar for each of the first 5 days?
Today, both Sam and Terri have no fish. They both go fishing each day. Sam catches 2 fish each day. Terri catches 4 fish each day. How many fish do they have after each of the five days? Make a graph of the number of fish.
Compare 3 x 2 + 5 and 3 x (2 + 5)
Which equation is true?
15 - 7 – 2 = 10
15 - (7 – 2) = 10
Parenthesis
Exponents
Multiplication
and
Division
Order of Operations
(left to right)
and
Subtraction
(left to right)
Expression or Equation?
Operation Comprehension
(P. 43)
Your objective in the beginning isn't necessarily to develop an area formula but to apply students' developing concept of multiplication to the area of rectangles
(Van de Walle, 2010)
Area, Perimeter, and Volume
Today we are going to find the area of a shape
Focus Question:
See if you can find another way to find
the area without counting each tile
8x4 = 32 and 8x3 = 24 32+24 = 56
Think back to the activity we just did
“A rectangular garden has as an area of 80 square feet. It is 5 feet wide. How long is the garden?” Here,specifying the area and the width, creates an unknown factor problem
Helping students understand the properties of operations for multiplication and division
MODELS!
Associative property
Commutative property
Distributive property
"What amount would be added to a quantity?"
"What factor would multiply one quantity?"
Consider two diving boards. One is 40 feet high and the other is 8 feet high.
In first and second grade students would compare these two values in an additive sense.
One is 32 feet higher than the other.
In fourth grade students start comparing the two value multiplicatively.
One is 5 times higher than the other.
How should you correctly interpret the statement "times more than"?
One person take three counters, and give another person three times more than you.
How many counters did you give your partner?
Ideally, the wording should be changed to "times as much"
We need to expose students to both interpretations, but let them know that typically the accepted interpretation would be just to use multiplication (e.g., 3 x 3)
Lines and Angles
(P.50)
Initially, an informal definition of a right angle, acute angle, and obtuse angle should be given.
1/360 degrees
An entire rotation is 360 degrees
Have students practice identifying a variety of different sized angles
"packing" volume is more difficult than iterating a unit to measure length and measuring by tiling.
eventually students learn to conceptalize a layer as a unit that itself is composed of units of rows, each row composed of individual cubes ....
Meaning must be attached to a unit
(Fractions)
Multiplication and Divison
(Fractions)
p. 34
Primary Objectives
3. how to utilize SEDOL's curricular tools to address Common Core State Standards
2. how to read SEDOL's Scope and Sequence
1. how to read SEDOL's Curriculum Framework
Understand:
Secondary Objectives
2. Introduce the instructional routine for the math block
1. Increase your math content knowledge so that you can analyze student development.
What are the Common Core Standards?
Designed with focus, coherence, and rigor in mind.
Narrows the scope of content in each grade
Domain
Cluster
Standard
8 Math Practice Standards
1. Make sense of problems and persevere in solving them.
2. Reason abstractly and quantitatively.
3. Construct viable arguments and critique the reasoning of others.
4. Model with mathematics.
5. Use appropriate tools strategically.
6. Attend to precision.
7. Look for and make use of structure.
8. Look for and express regularity in repeated reasoning
HOW
Math should be taught
Time and Money
P.45
Measurement
P.47
(p. 19 )
P. 28
P. 31
p. 56
P. 31
Be careful of cases involving 0 in division
Represent and Interpret Data
P. 53
Measure the length of your shoe in inches.
What length(s) were the most common?
How many more people had lengths of 7 than 6 1/4?
How many students had measurements larger than 7?
What is the total shoe length for the students who measured six and one-half?
What is the total number of inches for shoes measuring six and one-half and seven and one-half?
If we added all the shoe lengths above seven and one-fourth together what would we get?
What would be the total length of both shoes for ALL the people who had shoe lengths of six and one-half?
Instructional Strategies(purple box)
Key Idea: A fraction does not say anything about the size of the whole or the size of the parts. A fraction tells us ONLY about the RELATIONSHIP between the part and whole.
Use a variety of fractional models
Fractional parts are the building block for all fraction concepts
Much practice with dividing shapes and representing on a number line to build this type of number sense (MP6)
Promote reasoning with benchmark fractions: 0, 1/2, 1/1
Misconception
Misconception:
After several examples that 8/12 = 2/3, one would make an argument that each was divided or multiplied by 4
(
Through reasoning
- they haven't learned about fraction multiplication)
Again they don't know about fraction multiplication, but are beginning to discover algorithm
4/5 or 4/9 (same numerator)
4/7 or 5/7 (same denominator)
3/8 or 4/10 (different numerator & denominator)
4/6 or 7/12
WHich is Greater
NOT expected to find common multiple - will discover using models
so, let's test our number sense using benchmark fractions
5/3 or 7/4 (remember we are always going to work with
fractions greater than 1)
Always
taught simultaneously - this is just
a new way to write the number
1. Using the meaning of a decimal as a fraction generalizes to work with decimals in grade 5 that have more than two digits (thousandths).
2 Arguments
2. Visual fraction models don't generalize well to 5th grade.
That's why it's so necessary to to build the conceptual knowledge and view decimals as fractions when first introducing.
see if this activity is easier now
Calculators
Shape Reasoning
P. 62
Coordinate Graphing
P. 60
The PARCC Assessment
The Partnership for Assessment of Readiness for College and Careers
http://www.parcconline.org/parcc-assessment
" The Mathematical Practices will be taken seriously in curriculum and teaching if, and only if, they are taken seriously in testing. It can be expected, then, that the developers of the CCSS, and the States that collaborated in calling for the development of the CCSS, will work with the developers of assessments to ensure that the Mathematical Practices are taken seriously in testing.
Math "IPF"
Primary Objectives
Secondary Objectives
1. Increase your math content knowledge so that you can analyze student development.
2. Introduce the instructional routine for the math block
TRY A NUMBER TALK!
Closing Thoughts
8+6=(8+2=10+4=14)
13-4=(13-3=10-1=9)
6+7=(6+6=12+1+13)
8+4=12, one knows 12-8=4
8+3=11, 3+8=11(commutative)
& 2+6+4=2+10=12(associative)
"Nice Numbers"
2. laying a stronger
foundation
1.
3. questioning
1/8 + 4/5 =
2/5 + 1/2 =
This will eliminate one of the biggest misconceptions of
2/5 + 1/2 = 3/7 because 3/7 isn't even a half
1 2 3
find end (given start & elapsed)
find start (given end & elapsed)
Chapter 2 - p.13
5292 divided by 42
If one were to take an AngLeg and rotate it all the way around, what would be formed?
same perimeter different area:
6x2
4x4
1x12
4x3
2x6
same area different perimeter
same perimeter different area:
12 x 1
4 x 3
6 x 2
6 x 2
4 x 4
same area different perimeter:
Usually this is based off of
place value understanding
1 2 3
1 2 3
The goal is for students to see unit fractions
as the basic building blocks of fractions, in the
same sense that the number 1 is the basic building
block of the whole numbers; just as every whole
number is obtained by combining a sufficient number
of 1s, every fraction is obtained by combining a
sufficient number of unit fractions.
build fraction language by
2. know what is meant by "equal parts"
1. specifying the whole
3. 5/3 is the quantity you get when combining 5
parts together when the whole is divided into 3 equal
parts
**no need to introduce "proper" or "improper" fractions
denominator is the unit we are counting
numerator is the number of those units we have
Full transcript | 4,413 | 17,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-09 | longest | en | 0.917244 |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_20 | 1,709,592,436,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00272.warc.gz | 108,532,675 | 13,760 | # 2014 AMC 10B Problems/Problem 20
## Problem
For how many integers $x$ is the number $x^4-51x^2+50$ negative?
$\textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16$
## Solution 1
First, note that $50+1=51$, which motivates us to factor the polynomial as $(x^2-50)(x^2-1)$. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^2-50<0. Solving this inequality, we find $1. There are exactly $12$ integers $x$ that satisfy this inequality, $\pm \{2,3,4,5,6,7\}$.
Thus our answer is $\boxed{\textbf {(C) } 12}$.
## Solution 2
Since the $x^4-51x^2$ part of $x^4-51x^2+50$ has to be less than $-50$ (because we want $x^4-51x^2+50$ to be negative), we have the inequality $x^4-51x^2<-50 \rightarrow x^2(x^2-51) <-50$. $x^2$ has to be positive, so $(x^2-51)$ is negative. Then we have $x^2<51$. We know that if we find a positive number that works, it's parallel negative will work. Therefore, we just have to find how many positive numbers work, then multiply that by $2$. If we try $1$, we get $1^4-51(1)^4+50 = -50+50 = 0$, and $0$ therefore doesn't work. Test two on your own, and then proceed. Since two works, all numbers above $2$ that satisfy $x^2<51$ work, that is the set {${2,3,4,5,6,7}$}. That equates to $6$ numbers. Since each numbers' negative counterparts work, $6\cdot2=\boxed{\textbf{(C) }12}$.
## Solution 3 (Graph)
As with Solution $1$, note that the quartic factors to $(x^2-50)\cdot(x^2-1)$, which means that it has roots at $-5\sqrt{2}$, $-1$, $1$, and $5\sqrt{2}$. Now, because the original equation is of an even degree and has a positive leading coefficient, both ends of the graph point upwards, meaning that the graph dips below the $x$-axis between $-5\sqrt{2}$ and $-1$ as well as $1$ and $5\sqrt{2}$. $5\sqrt{2}$ is a bit more than $7$ ($1.4\cdot 5=7$) and therefore means that $-7,-6,-5,-4,-3,-2,2,3,4,5,6,7$ all give negative values.
## Solution 4
Let $x^{2}=u$. Then the expression becomes $u^{2}-51u+50$ which can be factored as $\left(u-1\right)\left(u-50\right)$. Since the expression is negative, one of $\left(u-1\right)$ and $\left(u-50\right)$ need to be negative. $u-1>u-50$, so $u-1>0$ and $u-50<0$, which yields the inequality $50>u>1$. Remember, since $u=x^{2}$ where $x$ is an integer, this means that $u$ is a perfect square between $1$ and $50$. There are $6$ values of $u$ that satisfy this constraint, namely $4$, $9$, $16$, $25$, $36$, and $49$. Solving each of these values for $x$ yields $12$ values (as $x$ can be negative or positive) $\Longrightarrow \boxed{\textbf {(C) } 12}$. ~JH. L | 945 | 2,707 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 69, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2024-10 | latest | en | 0.899945 |
https://boardgames.stackexchange.com/questions/53888/should-i-take-out-a-double-using-a-two-card-suit/53892 | 1,716,311,882,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058504.35/warc/CC-MAIN-20240521153045-20240521183045-00470.warc.gz | 112,012,190 | 39,092 | # Should I "take out" a double, using a two card suit?
In an earlier question, the consensus was that I should answer a takeout double of one heart with one spade (the cheaper bid) rather than two clubs. (I agree with that consensus.) The exact situation was: "Left hand opponent opened 1 heart. Partner doubled for takeout. Right hand opponent passed." I had something like (S) xxx (H) xx (D) Jxxx (C) Jxxx.
Suppose my major suits were reversed, so I had (S) xx (H) xxx (D) Jxxx (C) Jxxx? Would you still bid "one spade" with only two spades, or is "2 clubs" now the lesser evil?
Suppose one of my minors had five cards instead of four: e.g. (S) xxx (H) xx (D) Jxx (C) Jxxxx (three spades), or (S) xx (H) xx (D) Jxxx (C) Jxxxx (only two spades).
In the original question and answer, "one spade" was the preferred bid versus two clubs when I had one fewer spade than clubs. But the above variations hypothesize that I have at least two fewer spades than clubs. Put another way, where is the "tipping point" of spade-club distributions, where you would bid two clubs rather than one spade?
• Not really relevant here, but another consideration possibly worth keeping in mind is the projected effect our bid has on partner's future evaluation. In this precise situation partner is aware that we may have a bust with only 3. If it goes (1C)-double-pass and I'm to bid with 3334 hand with two jacks or less, I will bid 1D. The reasons are A) it keeps bidding low (see Forget's answer for the reason why we want that to happen), and B) it is less like to excite partner unduly should they have a strong hand. Jan 12, 2021 at 17:30
With `xxx xx Jxxx Jxxx`, the decision between 1S and 2C is very close. Any move that makes spades less attractive (eg holding only a doubleton) or clubs more attractive (eg a 5-card suit) breaks the "tie" in favor of 2C. Any of those three hands would be a clear 2C call for me.
• or even xxx xx Jxxx QJxx. We don't bid 1S with the original garbage hand because it's a good bid, we bid it because it keeps us on the 1 level, letting partner show their suit if they are double-and-bidding, giving us a runout if we need it, but mostly hoping the opponents will let us off the hook. It's really hard to double at the 1 level (more so now that the double of 1S is likely to be considered takeout); much easier to double a misfit at the 2 level - so they likely are going to rescue us. Jan 10, 2021 at 19:08
I would not be inclined to bid 1S with any of the hands mentioned, nor in most circumstances with a 2-card spade suit. With a minimum hand responding 1S on a three card suit might be better than 2C on four to the jack -- the decision is marginal. But bidding 1S with only 2 spades could easily find you in a 6- or even a 5-card fit, in a very bad place. I would absolutely bid a 5-card club suit over a 2-card spade suit, even 5 small cards. Four to the jack is not as good, particularly in a minor, but is still likely to work better than bidding a 2-card spade suit.
About the only time I would bid a 2-card suit in response to a takeout double would be if the opponents had bid 2 suits and I was 4-4 in those suits, but not strong enough to pass for penalties
Say !D - P - 1H - Dbl P
holding
Jx xxxx xxxx kxx
Passing 1H doubled could be a disaster, and 1S might be better than 2C on a 3-card suit I am certainly hoping that the opponents bid on and come to grief in a red suit at a higher level.
But holding (S) xx (H) xxx (D) Jxxx (C) Jxxx? I would bid 2C and hope that opps bid on or partner happens to have a club suit. I don't expect to do well, but anything else is likely to be worse. Sure you can find hands on which 1S is the killing bid. But not many with that bidding and that hand.
Two points not raised by the other answers are:
1. What's happening on the hand?
When partner Doubles 1H with our bust and 3-small Spade holding, expectation is that Partner is either:
• Holding 4 Spade cards and short Hearts, investigating our ability to compete, and will either pass or raise 1S depending on strength;
• A bit stronger intending a minimum NT bid to show a balanced 18-19 HCP with a Heart stopper; or
• A 1-suited hand of his own too strong for a simple overcall.
Our goal is in all cases to exit the auction as soon as possible, without ever giving opponents an opportunity to consider a penalty double that might be worth more than their own contract. To this end, it's imperative to both stay low, and avoid bidding a suit the opponents might hold 7 (or more) of. This is why the 1S response is acceptable on 3-small and a bust: Partner is either bidding on, which we can pass, or has a 4-card Spade himself. If Partner's intent is to show a 18-19 NT hand it is valuable to allow that to happen as 1 NT instead of as 2NT.
However when holding just 2-small the dynamics change. This could be going down 3 or even 4 against a mere part score by Opponents, especially with partner's hand on the table. Bidding a suit they hold discourages them from bidding, when we wish to make their continuation as easy as possible.
Hence I lean very much into bidding 1S on the xxx-xx-Jxxx-Jxxx hand but will never continue with a 1S call holding a 5-card minor of any quality, even 5 small. There is good reason to expect 2 long Club tricks in my hand after a 2C response, even with xxxxx, compared to zero tricks total in any other denomination.
Note that while Partner might have doubled on a three-suiter (short hearts) and only 3 Spade cards, such a call must guarantee better than minimum doubling values. A minimum holding and lacking Spades is better to wait, retaining the option of reopening later. | 1,474 | 5,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-22 | latest | en | 0.966273 |
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# EXCEL Formulas Bible
## Excel 2013 / 2016
1. SUM of Digits when cell Contains all Numbers ......................................................................................1
2. SUM of Digits when cell Contains Numbers and non Numbers both ..........................................1
3. A List is Unique or Not (Whether it has duplicates) ...........................................................................1
4. Count No. of Unique Values ............................................................................................................................. 1
5. Count No. of Unique Values Conditionally ...............................................................................................1
6. Find Last Day of the Month ............................................................................................................................. 2
7. Find First Day of the Month ............................................................................................................................ 2
8. Add Month to or Subtract Month from a Given Date ..........................................................................2
9. Add Year to or Subtract Year from a Given Date...................................................................................3
10. Last Day of the Month for a Given Date ................................................................................................3
11. Convert a Number to a Month Name .....................................................................................................3
12. Convert a Month Name to Number .........................................................................................................4
13. Convert a Number to Weekday Name ...................................................................................................4
14. Convert a Weekday Name to Number ...................................................................................................4
15. Converting Date to a Calendar Quarter.................................................................................................5
16. Converting Date to a Indian Financial Year Quarter ......................................................................5
17. Calculate Age from Given Birthday .........................................................................................................5
18. Number to Date Format Conversion ......................................................................................................5
19. Number to Time Format Conversion .....................................................................................................6
20. Number of Days in a Month ........................................................................................................................6
21. How to Know if a Year is a Leap Year ....................................................................................................6
22. Last Working Day of the Month If a Date is Given ...........................................................................6
23. First Working Day of the Month if a Date is Given ..........................................................................7
24. First Day of the Month for a Given Date ...............................................................................................7
25. How Many Mondays or any other Day of the Week between 2 Dates ...................................8
26. Date for Nth Day of the Year ......................................................................................................................8
27. Extract Date and Time from Date Timestamp ...................................................................................8
28. Financial Year Formula (e.g. 2015-16 or FY16) ...............................................................................8
29. First Working Day of the Year ...................................................................................................................9
30. Last Working Day of the Year ....................................................................................................................9
31. Convert from Excel Date (Gregorian Date) to Julian Date ........................................................ 10
32. Convert from Julian Dates to Excel (Gregorian) Dates ............................................................... 10
33. Convert a Number into Years and Months ....................................................................................... 10
34. Find the Next Week of the Day............................................................................................................... 11
35. Find the Previous Week of the Day ...................................................................................................... 12
36. Count Cells Starting (or Ending) with a particular String......................................................... 12
37. Count No. of Cells Having Numbers Only .......................................................................................... 13
38. Count No. of Cells which are containing only Characters ......................................................... 13
39. Number of Characters in a String without considering blanks .............................................. 13
40. Number of times a character appears in a string .......................................................................... 13
41. Count Non Numbers in a String ............................................................................................................. 13
42. Count Numbers in a String ....................................................................................................................... 13
43. Count only Alphabets in a String........................................................................................................... 14
44. Most Frequently Occurring Value in a Range ................................................................................. 14
45. COUNTIF on Filtered List .......................................................................................................................... 14
46. SUMIF on Filtered List................................................................................................................................ 15
47. Extract First Name from Full Name ..................................................................................................... 15
48. Extract Last Name from Full Name ...................................................................................................... 15
49. Extract the Initial of Middle Name ....................................................................................................... 15
50. Extract Middle Name from Full Name ................................................................................................ 15
51. Remove Middle Name in Full Name .................................................................................................... 15
52. Extract Integer and Decimal Portion of a Number ....................................................................... 15
53. Maximum Times a Particular Entry Appears Consecutively ................................................... 16
54. Get File Name through Formula ............................................................................................................ 17
55. Get Workbook Name through Formula ............................................................................................. 17
56. Get Sheet Name through Formula ........................................................................................................ 17
57. Get Workbook's Directory from Formula ......................................................................................... 17
58. Perform Multi Column VLOOKUP ......................................................................................................... 18
59. VLOOKUP from Right to Left ................................................................................................................... 19
60. Case Sensitive VLOOKUP .......................................................................................................................... 19
61. Rank within the Groups ............................................................................................................................. 20
62. Remove Alphabets from a String .......................................................................................................... 20
63. Remove numbers from string................................................................................................................. 21
64. Roman Representation of Numbers .................................................................................................... 21
65. Sum Bottom N Values in a Range .......................................................................................................... 21
66. Sum Every Nth Row..................................................................................................................................... 22
67. We have AVERAGEIF. What about MEDIANIF and MODEIF? ................................................. 22
68. Calculate Geometric Mean by Ignoring 0 and Negative Values .............................................. 23
69. Financial Function - Calculate EMI....................................................................................................... 23
70. Financial Function - Calculate Interest Part of an EMI ............................................................... 24
71. Financial Function - Calculate Principal Part of an EMI ............................................................ 26
72. Financial Function - Calculate Number of EMIs to Pay Up a Loan ........................................ 27
73. Financial Function - Calculate Interest Rate ................................................................................... 28
74. Financial Function – Calculate Compounded Interest ................................................................ 30
75. Financial Function – Calculate Effective Interest .......................................................................... 31
76. Abbreviate Given Names........................................................................................................................... 32
77. Get Column Name for a Column Number .......................................................................................... 33
78. Get Column Range for a Column Number ......................................................................................... 33
79. Find the nth Largest Number when there are duplicates ......................................................... 33
80. COUNTIF for non-contiguous range .................................................................................................... 34
81. Count the Number of Words in a Cell / Range ............................................................................... 35
82. Numerology Sum of the Digits aka Sum the Digits till the result is a single digit.......... 35
83. Generate Sequential Numbers and Repeat them .......................................................................... 36
84. Repeat a Number and Increment and Repeat.... ............................................................................ 36
85. Generate Non Repeating Random Numbers through Formula .............................................. 37
86. Extract User Name from an E Mail ID ................................................................................................. 37
87. Extract Domain Name from an E Mail ID .......................................................................................... 37
88. Location of First Number in a String ................................................................................................... 38
89. Location of Last Number in a String .................................................................................................... 38
90. Find the Value of First Non Blank Cell in a Range......................................................................... 38
91. Find First Numeric Value in a Range ................................................................................................... 38
92. Find Last Numeric Value in a Range .................................................................................................... 38
93. Find First non Numeric Value in a Range ......................................................................................... 38
94. Find Last non Numeric Value in a Range .......................................................................................... 38
95. Find Last Used Value in a Range ........................................................................................................... 38
96. MAXIF ................................................................................................................................................................. 38
97. MINIF .................................................................................................................................................................. 39
98. Generate a Conditional List ..................................................................................................................... 39
99. Generate a Unique List out of Duplicate Entries ........................................................................... 40
Excel Formulas Bible
## 1. SUM of Digits when cell Contains all Numbers
If you cell contains only numbers like A1:= 7654045, then following formula can be used to
find sum of digits
=SUMPRODUCT(--MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1))
## 2. SUM of Digits when cell Contains Numbers and non
Numbers both
If you cell contains non numbers apart from numbers like A1:= 76\$5a4b045%d, then
following formula can be used to find sum of digits
=SUMPRODUCT((LEN(A1)-LEN(SUBSTITUTE(A1,ROW(1:9),"")))*ROW(1:9))
The above formula can be used even if contains all numbers as well.
## 3. A List is Unique or Not (Whether it has duplicates)
Assuming, your list is in A1 to A1000. Use following formula to know if list is unique.
=MAX(FREQUENCY(A1:A1000,A1:A1000))
=MAX(INDEX(COUNTIF(A1:A1000,A1:A1000),,))
## 4. Count No. of Unique Values
Use following formula to count no. of unique values -
=SUMPRODUCT((A1:A100<>"")/COUNTIF(A1:A100,A1:A100&""))
## 5. Count No. of Unique Values Conditionally
If you have data like below and you want to find the unique count for Region = “A”, then you
can use below Array formula –
=SUM(IF(FREQUENCY(IF(A2:A20<>"",IF(A2:A20="A",MATCH(B2:B20,B2:B20,0))),ROW(A
2:A20)-ROW(A2)+1),1))
If you have more number of conditions, the same can be built after A2:A20 = “A”.
Note - Array Formula is not entered by pressing ENTER after entering your formula but by
pressing CTRL+SHIFT+ENTER. If you are copying and pasting this formula, take F2 after
pasting and CTRL+SHIFT+ENTER. This will put { } brackets around the formula which you
can see in Formula Bar. If you edit again, you will have to do CTRL+SHIFT+ENTER again.
Don't put { } manually.
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## 6. Find Last Day of the Month
Suppose, you have a date in the cell A1≔ 14-Aug-16, then formula for finding last day of the
month is
=EOMONTH(A1,0)
## 7. Find First Day of the Month
Suppose, you have a date in the cell A1≔ 14-Aug-16, then formula for finding first day of
the month is
=A1-DAY(A1)+1
OR
=EOMONTH(A1,-1)+1
## 8. Add Month to or Subtract Month from a Given Date
Very often, you will have business problems where you have to add or subtract month
from a given date. One scenario is calculation for EMI Date.
Say, you have a date of 10/22/14 (MM/DD/YY) in A1 and you want to add number of
months which is contained in Cell B1.
=EDATE(A1,B1)
## [Secondary formula =DATE(YEAR(A1),MONTH(A1)+B1,DAY(A1)) ]
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=EDATE(A1,-B1)
## 9. Add Year to or Subtract Year from a Given Date
In many business problems, you might encounter situations where you will need to add or
subtract years from a given date.
## If you want to add Years to a given date, formulas would be -
=EDATE(A1,12*B1)
=DATE(YEAR(A1)+B1,MONTH(A1),DAY(A1))
## If you want to subtract Years from a given date, formulas would be -
=EDATE(A1,-12*B1)
=DATE(YEAR(A1)-B1,MONTH(A1),DAY(A1))
## 10. Last Day of the Month for a Given Date
Suppose, you are given a date say 10/22/14 (MM/DD/YY) and we want to have the last
date of the month for the given date. Hence, you needs an answer of 10/31/14. The formulas
to be used in this case -
=EOMONTH(A1,0)
=DATE(YEAR(A1),MONTH(A1)+1,0)
=DATE(YEAR(A1),MONTH(A1)+1,1)-1
## 11. Convert a Number to a Month Name
Use below formula to generate named 3 lettered month like Jan, Feb....Dec
=TEXT(A1*30,"mmm")
Replace "mmm" with "mmmm" to generate full name of the month like January,
February....December in any of the formulas in this post.
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## 12. Convert a Month Name to Number
Say Cell A1 contains the string January, February….December (or Jan. Feb…..Dec) and you
want to show 1, 2……12
=MONTH("1"&LEFT(A1,3))
## 13. Convert a Number to Weekday Name
Suppose you want to return 1 = Sunday, 2 = Monday…..7 = Saturday
=TEXT(A1&"Jan2017","dddd")
## To show only 3 characters of the Weekday Name
=TEXT(A1&"Jan2017","ddd")
You can add a number to A1 if you want to show some other Weekday Name
## Say, if you want to show 1 = Monday, 2 = Tuesday…….7 = Sunday, just add 1 to A1
=TEXT(A1+1&"Jan2017","dddd")
## Say, if you want to show 1 = Friday, 2 = Saturday…….7 = Thursday, just add 5 to A1
=TEXT(A1+5&"Jan2017","dddd")
## 14. Convert a Weekday Name to Number
Say Cell A1 contains the string Sunday, Monday….Saturday (or Sun, Mon…..Sat) and you
want to show 1, 2…..7
Suppose Cell A1 contains weekday names like Sunday, Monday.....(or Sun, Mon...), then
following formula can be used to return the numbers. Sunday will be 1 and Saturday will be
7.
=ROUND(SEARCH(LEFT(A1,2),"SuMoTuWeThFrSa")/2,0)
=MATCH(LEFT(A1,2),{"Su","Mo","Tu","We","Th","Fr","Sa"},0)
If we want to return some other number to weekdays, then formula can be tweaked
accordingly. For example, to make Mon = 1 and Sun = 7
=ROUND(SEARCH(LEFT(A1,2),"MoTuWeThFrSaSu")/2,0)
=MATCH(LEFT(A1,2),{"Mo","Tu","We","Th","Fr","Sa","Su"},0)
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## 15. Converting Date to a Calendar Quarter
Assuming date is in Cell A1. You want to convert it into a quarter (1, 2, 3 & 4). Jan to Mar is
1, Apr to Jun is 2, Jul to Sep is 3 and Oct to Dec is 4.
=CEILING(MONTH(A1)/3,1)
OR
= ROUNDUP(MONTH(A1)/3,0)
OR
=CHOOSE(MONTH(A1),1,1,1,2,2,2,3,3,3, 4,4,4)
## 16. Converting Date to a Indian Financial Year Quarter
Assuming date is in Cell A1. You want to convert it into a Indian Financial Year Quarter. Jan
to Mar is 4, Apr to Jun is 1, Jul to Sep is 2 and Oct to Dec is 3.
=CEILING(MONTH(A1)/3,1)+IF(MONTH(A1)<=3,3,-1)
OR
=ROUNDUP(MONTH(A1)/3,0)+IF(MONTH(A1)<=3,3,-1)
OR
=CHOOSE(MONTH(A1),4,4,4,1,1,1,2,2,2,3,3,3)
## 17. Calculate Age from Given Birthday
=DATEDIF(A1,TODAY(),"y")&" Years "&DATEDIF(A1,TODAY(),"ym")&" Months
"&DATEDIF(A1,TODAY(),"md")&" Days"
## 18. Number to Date Format Conversion
If you have numbers like 010216 and you want to convert this to date format, then the
following formula can be used
## Note – Minimum 5 digits are needed for above formula to work
If you have numbers like 01022016 and you want to convert this to date format, then the
following formula can be used
## =--TEXT(A1,"00\/00\/0000") for 4 digits year
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## 19. Number to Time Format Conversion
If you have numbers like 1215 and you want to convert this to hh:mm format, then the
following formula can be used
=--TEXT(A1,"00\:00")
## To convert to hh:mm:ss format
=--TEXT(A1,"00\:00\:00")
## 20. Number of Days in a Month
Suppose, you have been given a date say 15-Nov-14 and you have to determine how many
days this particular month contains.
The formula which you need to use in the above case would be
=DAY(EOMONTH(A1,0))
Explanation - EOMONTH(A1,0) gives the last date of the month and DAY function extract
that particular Day from the last date of the month.
## 21. How to Know if a Year is a Leap Year
Let's say that A1 contains the year. To know whether it is a Leap Year or not, use following
formula -
=MONTH(DATE(A1,2,29))=2
TRUE means that it is Leap Year and FALSE means that this is not a Leap Year.
## 22. Last Working Day of the Month If a Date is Given
If A1 holds a date, the formula for calculating last Working Day of the month would be
=WORKDAY(EOMONTH(A1,0)+1,-1)
The above formula assumes that your weekends are Saturday and Sunday.
But, if your weekends are different (e.g. in gulf countries), you can use following formula -
=WORKDAY.INTL(EOMONTH(A1,0)+1,-1,"0000110")
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Where 0000110 is a 7 character string, 1 represents a weekend and 0 is a working day. First
digit is Monday and last digit is Sunday. The above example is for Gulf countries where
Friday and Saturday are weekends.
You also have option to give a range which has holidays. In that case, your formula would
become
=WORKDAY(EOMONTH(A1,0)+1,-1,D1:D10)
=WORKDAY.INTL(EOMONTH(A1,0)+1,-1,"0000110",D1:D10)
## 23. First Working Day of the Month if a Date is Given
If A1 contains a date, then formula for First Working Day of the month would be
=WORKDAY(EOMONTH(A1,-1),1)
The above formula assumes that your weekends are Saturday and Sunday.
But, if your weekends are different (e.g. in gulf countries), you can use following formula -
=WORKDAY.INTL(EOMONTH(A1,-1),1,"0000110")
Where 0000110 is a 7 character string, 1 represents a weekend and 0 is a working day. First
digit is Monday and last digit is Sunday. The above example is for Gulf countries where
Friday and Saturday are weekends.
You also have option to give a range which has holidays. In that case, your formula would
become
=WORKDAY(EOMONTH(A1,-1),1,D1:D10)
=WORKDAY.INTL(EOMONTH(A1,-1),1,"0000110",D1:D10)
## 24. First Day of the Month for a Given Date
Suppose you have been given a date say 10/22/14 (MM/DD/YY) and you want to calculate
the first day of the Current Month. Hence, you want to achieve a result of 10/1/2014
(MM/DD/YY).
## The formulas to be used -
=DATE(YEAR(A1),MONTH(A1),1)
=A1-DAY(A1)+1
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=EOMONTH(A1,-1)+1
## 25. How Many Mondays or any other Day of the Week
between 2 Dates
Suppose A1 = 23-Jan-16 and A2 = 10-Nov-16. To find number of Mondays between these
two dates
=SUMPRODUCT(--(TEXT(ROW(INDIRECT(A1&":"&A2)),"ddd")="Mon"))
“Mon” can be replaced with any other day of the week as per need.
## 26. Date for Nth Day of the Year
Suppose A1 contains the Year and you are asked to find 69th day of the year which is
contained in A2. Then formula for finding Nth day of the year would be
=DATE(A1,1,1)+A2-1
## 27. Extract Date and Time from Date Timestamp
Suppose you have a date timestamp value in cell A1
A1 = 06/14/15 10:15 PM
## And you want to extract date and time out of this.
To extract date, use following formula and format the result cell as date
= INT(A1)
To extract time, use following formula and format the result cell as time
= MOD(A1,1)
## 28. Financial Year Formula (e.g. 2015-16 or FY16)
A good number of countries don't follow calendar year as the financial year. For example,
India's financial year start is 1-Apr and finishes on 31-Mar. Hence, currently (20-Feb-16),
the financial year is 2015-16 (It is also written as FY16). On 1-Apr-16, it will become 2016-
17 (It is also written as FY17).
Now if a date is given, then following formula can be used to derive 2015-16 kind of result.
=YEAR(A1)-(MONTH(A1)<=3)&"-"&YEAR(A1)+(MONTH(A1)>3)
## To generate FY16 kind of result, following formula can be used
="FY"&RIGHT(YEAR(A1)+(MONTH(A1)>3),2)
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## 29. First Working Day of the Year
If a year is given in A1 say 2016, below formula can be used to know the first working day
of the year (format the result as date)
=WORKDAY(EOMONTH("1JAN"&A1,-1),1)
The above formula assumes that your weekends are Saturday and Sunday.
But, if your weekends are different (e.g. in gulf countries), you can use following formula -
=WORKDAY.INTL(EOMONTH("1JAN"&A1,-1),1,"0000110")
Where 0000110 is a 7 character string, 1 represents a weekend and 0 is a working day. First
digit is Monday and last digit is Sunday. The above example is for Gulf countries where
Friday and Saturday are weekends.
You also have option to give a range which has holidays. In that case, your formula would
become
=WORKDAY(EOMONTH("1JAN"&A1,-1),1,D1:D10)
=WORKDAY.INTL(EOMONTH("1JAN"&A1,-1),1,"0000110",D1:D10)
## 30. Last Working Day of the Year
If a year is given in A1 say 2016, below formula can be used to know the last working day
of the year (format the result as date)
=WORKDAY("1JAN"&A1+1,-1)
The above formula assumes that your weekends are Saturday and Sunday.
But, if your weekends are different (e.g. in gulf countries), you can use following formula -
=WORKDAY.INTL("1JAN"&A1+1,-1,"0000110")
Where 0000110 is a 7 character string, 1 represents a weekend and 0 is a working day. First
digit is Monday and last digit is Sunday. The above example is for Gulf countries where
Friday and Saturday are weekends.
You also have option to give a range which has holidays. In that case, your formula would
become
=WORKDAY("1JAN"&A1+1,-1,D1:D10)
=WORKDAY.INTL("1JAN"&A1+1,-1,"0000110",D1:D10)
Where range D1:D10 contains the list of holidays.
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## 31. Convert from Excel Date (Gregorian Date) to Julian
Date
Q. First what is a Julian Date?
A. A Julian date has either 7 digits or 5 digits date and these are generally used in old IT
legacy systems.
7 Digits - YYYYDDD - 2016092 (This is 1-Apr-2016. 92 means that this is 92nd day from 1-
Jan in that year)
5 Digits - YYDDD - 16092
## Q. What formulas to use to convert Excel Dates to Julian Dates?
A. For 7 Digits, use following formula
=TEXT(A1,"yyyy")&TEXT(A1-("1JAN"&YEAR(A1))+1,"000")
## For 5 Digits, use following formula
=TEXT(A1,"yy")&TEXT(A1-("1JAN"&YEAR(A1))+1,"000")
## 32. Convert from Julian Dates to Excel (Gregorian) Dates
For 7 Digits Julian Dates, following formula should be used
=DATE(LEFT(A1,4),1,RIGHT(A1,3))
For 5 Digits Julian Dates, following formula should be used depending upon which century
(Note - Julian dates are most likely to fall into 20th Century)
21st Century
=DATE(20&LEFT(A1,2),1,RIGHT(A1,3))
20th Century
=DATE(19&LEFT(A1,2),1,RIGHT(A1,3))
Note - 19 or 20 can be replaced with some IF condition to put in right 19 or 20 depending
upon the year. For example, year 82 is more likely to be in 20th century where year 15 is
more likely to be in 21st century.
## 33. Convert a Number into Years and Months
Suppose, you have been given a number into cell A1 say 26 and you want to display it as 2
Years and 4 Months, you can use following formula -
## =INT(A1/12)&" Years and "&MOD(A1,12)&" Months"
Now, an user can become more demanding and he can say that if month is less than 12, then
Years should not be displayed. For example, he might say that 8 should be converted to 8
Months and it should not be shown as 0 Years and 8 Months.
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## Now 8 will be displayed as 8 Months only not as 0 Years and 8 Months.
Now, user can ask more. He can say when I give 12, it displays as 1 Years and 0 Months and
he simply wants to see 1 Years only. And for 36, he wants to see only 3 Years not 3 Years 0
Months. In this case, formula will have to be tweaked more. Now, the formula becomes -
## =IF(INT(A1/12)>0,INT(A1/12)&" Years ","")&IF(MOD(A1,12)=0,"",MOD(A1,12)&"
Months")
Now an user can come and can ask for one last thing. He can say that if this is 1 Year or 1
Month, it should not be displayed as Years or Months as 1 is not plural. Hence, 25 should be
displayed as 2 Years and 1 Month not as 2 Years and 1 Months. Hence, 18 should not be
displayed as 1 Years and 6 Months but as 1 Year and 6 Months. Similarly 13 should be
displayed as 1 Year and 1 Month not as 1 Years and 1 Months.
## =IF(INT(A1/12)>0,INT(A1/12)&" Year"&IF(INT(A1/12)>1,"s","")&" and
","")&MOD(A1,12)&" Month"&IF(MOD(A1,12)>1,"s","")
## 34. Find the Next Week of the Day
There are 2 scenarios in this case. For example, if today’s date is 2-Jan-17 (Monday) and I
try to find the next Monday, I can get either 2-Jan-17 or 9-Jan-17 as per need. For Tuesday
to Sunday, it is not a problem as they come after 2-Jan-17 only.
Case 1 - If the Day falls on the same date, then that very date (Hence, in case of 2-Jan-
17, next Monday would be 2-Jan-17 only)
## Next Mon =CEILING(\$A\$1-2,7)+2
Next Tue =CEILING(\$A\$1-3,7)+3
Next Wed =CEILING(\$A\$1-4,7)+4
Next Thu =CEILING(\$A\$1-5,7)+5
Next Fri =CEILING(\$A\$1-6,7)+6
Next Sat =CEILING(\$A\$1-7,7)+7
Next Sun =CEILING(\$A\$1-8,7)+8
Case 2 - If the Day falls on the same date, then next date (Hence, in case of 2-Jan-17, next
Monday would be 9-Jan-17 only)
## Next Mon =CEILING(\$A\$1-1,7)+2
Next Tue =CEILING(\$A\$1-2,7)+3
Next Wed =CEILING(\$A\$1-3,7)+4
Next Thu =CEILING(\$A\$1-4,7)+5
Next Fri =CEILING(\$A\$1-5,7)+6
Next Sat =CEILING(\$A\$1-6,7)+7
Next Sun =CEILING(\$A\$1-7,7)+8
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## 35. Find the Previous Week of the Day
There are 2 scenarios in this case. For example, if today’s date is 2-Jan-17 (Monday) and I
try to find the previous Monday, I can get either 2-Jan-17 or 26-Dec-16 as per need. For
Tuesday to Sunday, it is not a problem as they come prior to 2-Jan-17 only.
Case 1 - If the Day falls on the same date, then that very date (Hence, in case of 2-Jan-
17, previous Monday would be 2-Jan-17 only)
## Previous Mon =CEILING(\$A\$1-8,7)+2
Previous Tue =CEILING(\$A\$1-9,7)+3
Previous Wed =CEILING(\$A\$1-10,7)+4
Previous Thu =CEILING(\$A\$1-11,7)+5
Previous Fri =CEILING(\$A\$1-12,7)+6
Previous Sat =CEILING(\$A\$1-13,7)+7
Previous Sun =CEILING(\$A\$1-14,7)+8
Case 2 - If the Day falls on the same date, then previous date (Hence, in case of 2-Jan-
17, previous Monday would be 26-Dec-16 only)
## Previous Mon =CEILING(\$A\$1-9,7)+2
Previous Tue =CEILING(\$A\$1-10,7)+3
Previous Wed =CEILING(\$A\$1-11,7)+4
Previous Thu =CEILING(\$A\$1-12,7)+5
Previous Fri =CEILING(\$A\$1-13,7)+6
Previous Sat =CEILING(\$A\$1-14,7)+7
Previous Sun =CEILING(\$A\$1-15,7)+8
## 36. Count Cells Starting (or Ending) with a particular
String
1. Say you want to count all cells starting with C
=COUNTIF(A1:A10,"c*")
## Suppose you want to find all cells starting with Excel.
=COUNTIF(A1:A10,"excel*")
2. For ending
=COUNTIF(A1:A10,"*c")
## Suppose you want to find all cells starting with Excel.
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=COUNTIF(A1:A10,"*excel")
## 37. Count No. of Cells Having Numbers Only
COUNT function counts only those cells which are having numbers.
=COUNT(A1:A10)
## 38. Count No. of Cells which are containing only
Characters
Hence, if your cell is having a number 2.23, it will not be counted as it is a number.
## Use below formula considering your range is A1:A10
=COUNTIF(A1:A10,"*")
## 39. Number of Characters in a String without considering
blanks
Say, you have a string like Vijay A. Verma and I need to know how many characters it has.
In this case, it has 12 including decimal and leaving blanks aside.
## Use below formula for the same -
=LEN(SUBSTITUTE(A1," ",""))
## 40. Number of times a character appears in a string
Suppose you want to count the number of times, character “a” appears in a string
=LEN(A1)-LEN(SUBSTITUTE(LOWER(A1),"a",""))
## 41. Count Non Numbers in a String
Suppose you have a string "abc123def45cd" and you want to count non numbers in this.
## If your string is in A1, use following formula in A1
=IF(LEN(TRIM(A1))=0,0,SUMPRODUCT(--NOT(ISNUMBER((--
MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1))))))
## 42. Count Numbers in a String
Suppose you have a string "abc123def43cd" and you want to count numbers in this.
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## If your string is in A1, use following formula -
=SUMPRODUCT(LEN(A1)-LEN(SUBSTITUTE(A1,ROW(1:10)-1,"")))
OR
=SUMPRODUCT(--ISNUMBER((--MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1))))
## 43. Count only Alphabets in a String
Suppose you have a string "Ab?gh123def%h*" and you want to count only Aphabets.
## Suppose your string is in A1, put following formula for this.
=SUMPRODUCT(LEN(A1)-
LEN(SUBSTITUTE(UPPER(A1),CHAR(ROW(INDIRECT("65:90"))),"")))
OR
=SUMPRODUCT(--(ABS(77.5-
CODE(MID(UPPER(A1),ROW(INDIRECT("A1:A"&LEN(A1))),1)))<13))
## 44. Most Frequently Occurring Value in a Range
Assuming, your range is A1:A10, enter the below formula as Array Formula i.e. not by
pressing ENTER after entering your formula but by pressing CTRL+SHIFT+ENTER. This will
put { } brackets around the formula which you can see in Formula Bar. If you edit again, you
will have to do CTRL+SHIFT+ENTER again. Don't put { } manually.
=INDEX(A1:A10,MATCH(MAX(COUNTIF(A1:A10,A1:A10)),COUNTIF(A1:A10,A1:A10),0))
## The non-Array version of above formula
=INDEX(A1:A10,MATCH(MAX(INDEX(COUNTIF(A1:A10,A1:A10),,)),INDEX(COUNTIF(A1:
A10,A1:A10),,),0))
## 45. COUNTIF on Filtered List
You can use SUBTOTAL to perform COUNT on a filtered list but COUNTIF can not be done
on a filtered list. Below formula can be used to perform COUNTIF on a filtered list
=SUMPRODUCT(SUBTOTAL(3,OFFSET(B2,ROW(B2:B20)-ROW(B2),))*(B2:B20>14))
## Here B2:B20>14 is like a criterion in COUNTIF (=COUNTIF(B2:B20,">14"))
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## 46. SUMIF on Filtered List
You can use SUBTOTAL to perform SUM on a filtered list but SUMIF can not be done on a
filtered list. Below formula can be used to perform SUMIF on a filtered list
=SUMPRODUCT(SUBTOTAL(9,OFFSET(B2,ROW(B2:B20)-ROW(B2),))*(B2:B20>14))
## 47. Extract First Name from Full Name
=LEFT(A1,FIND(" ",A1&" ")-1)
## 48. Extract Last Name from Full Name
=TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",LEN(A1))),LEN(A1)))
## 49. Extract the Initial of Middle Name
Suppose, you have a name John Doe Smith and you want to show D as middle initial.
Assuming, your data is in A1, you may use following formula
## =IF(COUNTIF(A1,"* * *"),MID(A1,FIND(" ",A1)+1,1),"")
If name is of 2 or 1 words, the result will be blank. This works on 3 words name only as
middle can be decided only for 3 words name.
## 50. Extract Middle Name from Full Name
=IF(COUNTIF(A1,"* * *"),MID(A1,FIND(" ",A1)+1,FIND(" ",A1,FIND(" ",A1)+1)-(FIND("
",A1)+1)),"")
## =IF(COUNTIF(A1,"* * *"),TRIM(MID(SUBSTITUTE(A1," ",REPT(" ",LEN(A1)),2),FIND("
",A1)+1,LEN(A1))),"")
## =IF(COUNTIF(A1,"* * *"),LEFT(REPLACE(A1,1,FIND(" ",A1),""),FIND("
",REPLACE(A1,1,FIND(" ",A1),""))-1))
## 51. Remove Middle Name in Full Name
=IF(COUNTIF(A1,"* * *"),LEFT(A1,FIND(" ",A1&" "))&TRIM(RIGHT(SUBSTITUTE(A1,"
",REPT(" ",LEN(A1))),LEN(A1))),"")
",A1),""),"")
## 52. Extract Integer and Decimal Portion of a Number
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=INT(A1)
=TRUNC(A1)
## Positive value in A1 - If A1 contains 84.65, then answer would be 84.
Negative value in A1 - If A1 contains -24.39, then answer would be -24.
If you want only +ve value whether value in A1 is -ve or +ve, the formula can have many
variants.
=INT(A1)*SIGN(A1) OR =TRUNC(A1)*SIGN(A1)
=INT(ABS(A1)) OR =TRUNC(ABS(A1))
=ABS(INT(A1)) OR = ABS(TRUNC(A1))
## To extract Decimal portion -
=MOD(ABS(A1),1)
=ABS(A1)-INT(ABS(A1))
## Positive value in A1 - If A1 contains 84.65, then answer would be 0.65.
Negative value in A1 - If A1 contains -24.39, then answer would be 0.39.
## 53. Maximum Times a Particular Entry Appears
Consecutively
Suppose, we want to count maximum times “A” appears consecutively,
## You may use following Array formula -
=MAX(FREQUENCY(IF(A2:A20="A",ROW(A2:A20)),IF(A2:A20<>"A",ROW(A2:A20))))
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Note - Array Formula is not entered by pressing ENTER after entering your formula but by
pressing CTRL+SHIFT+ENTER. If you are copying and pasting this formula, take F2 after
pasting and CTRL+SHIFT+ENTER. This will put { } brackets around the formula which you
can see in Formula Bar. If you edit again, you will have to do CTRL+SHIFT+ENTER again.
Don't put { } manually.
## 54. Get File Name through Formula
Before getting this, make sure that you file has been saved at least once as this formula is
dependent upon the file path name which can be pulled out by CELL function only if file has
been saved at least once.
=CELL("filename",\$A\$1)
## 55. Get Workbook Name through Formula
Before getting this, make sure that you file has been saved at least once as this formula is
dependent upon the file path name which can be pulled out by CELL function only if file has
been saved at least once.
=REPLACE(LEFT(CELL("filename",\$A\$1),FIND("]",CELL("filename",\$A\$1))-
1),1,FIND("[",CELL("filename",\$A\$1)),"")
## 56. Get Sheet Name through Formula
Before getting this, make sure that you file has been saved at least once as this formula is
dependent upon the file path name which can be pulled out by CELL function only if file has
been saved at least once.
## Use following formula -
=REPLACE(CELL("filename",A1),1,FIND("]",CELL("filename",A1)),"")
Make sure that A1 is used in the formula. If it is not used, it will extract sheet name for the
last active sheet which may not be one which we want.
If you want the sheet name for last active sheet only, then formula would become
=REPLACE(CELL("filename"),1,FIND("]",CELL("filename")),"")
## 57. Get Workbook's Directory from Formula
Before getting this, make sure that you file has been saved at least once as this formula is
dependent upon the file path name which can be pulled out by CELL function only if file has
been saved at least once.
If your workbook is located in say C:\Excel\MyDocs, the formula to retrieve the directory
for this would be
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=LEFT(CELL("filename",A1),FIND("[",CELL("filename",A1))-2)
## 58. Perform Multi Column VLOOKUP
You know VLOOKUP, one of the most loved function of Excel. The syntax is
VLOOKUP(lookup_value,table_array,col_index_num,range_lookup)
## Here look_value can be a single value not multiple values.
Now, you are having a situation where you want to do vlookup with more than 1 values. For
the purpose of illustrating the concept, let's say we have 2 values to be looked up.
Below is your lookup table and you want to look up for Emp - H and Gender - F for Age.
=INDEX(C2:C12,MATCH(1,INDEX(--((A2:A12=F2)*(B2:B12=G2)*(ROW(A2:A12)-
ROW(A2)+1)<>0),,),0))
Concatenation Approach
=INDEX(C2:C10,MATCH(F2&"@@@"&G2,INDEX(A2:A10&"@@@"&B2:B10,,),0))
@@@ can be replaced by any characters which should not be part of those columns.
## By concatenation, you can have as many columns as possible.
CAUTION - Result of entire concatenation should not be having length more than 255.
Hence, F2&"@@@"&G2 should not have more than 255 characters.
## Another alternative is to use below Array formula -
=INDEX(C2:C12,MATCH(1,--NOT(ISLOGICAL(IF(A2:A12=F2,IF(B2:B12=G2,C2:C12)))),0))
Note - Array Formula is not entered by pressing ENTER after entering your formula but by
pressing CTRL+SHIFT+ENTER. If you are copying and pasting this formula, take F2 after
pasting and CTRL+SHIFT+ENTER. This will put { } brackets around the formula which you
can see in Formula Bar. If you edit again, you will have to do CTRL+SHIFT+ENTER again.
Don't put { } manually.
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## 59. VLOOKUP from Right to Left
VLOOKUP always looks up from Left to Right. Hence, in the below table, I can find Date of
Birth of Naomi by giving following formula –
=VLOOKUP("Naomi",B:D,3,0)
But, If I have to find Emp ID corresponding to Naomi, I can not do it through VLOOKUP
formula. To perform VLOOKUP from Right to Left, you will have to use INDEX / MATCH
combination. Hence, you will have to use following formula –
=INDEX(A:A,MATCH("Naomi",B:B,0))
## 60. Case Sensitive VLOOKUP
Suppose your have data like below table and you want to do a case sensitive VLOOKUP
If perform a regular VLOOKUP on SARA, I would get the answer 4300. But in a case sensitive
VLOOKUP, answer should be 3200. You may use below formula for Case Sensitive VLOOKUP
-
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=INDEX(B2:B10,MATCH(TRUE,INDEX(EXACT("SARA",A2:A10),,),FALSE))
## 61. Rank within the Groups
Suppose your have data like below table and you want to know rank of students.
## You will simple put following formula in D2
=RANK(C2,C2:C100)
But what if you are asked to produce rank of students within each school. Hence, every
Put following formula in D2 for that case for Descending order ranking. (For ascending
order, replace ">" with "<" without quote marks)
=SUMPRODUCT((B\$2:B\$100=B2)*(C\$2:C\$100>C2))+1
OR
=COUNTIFS(B\$2:B\$100,B2,C\$2:C\$100,">"&C2)+1
## 62. Remove Alphabets from a String
If your string is in cell A1, use following formula to remove all alphabets from a string
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=SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(
SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(
SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(
SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(
SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(
SUBSTITUTE(LOWER(A1),"a",""),"b",""),"c",""),"d",""),"e",""),"f",""),
"g",""),"h",""),"i",""),"j",""),"k",""),"l",""),"m",""),"n",""),"o",""),
"p",""),"q",""),"r",""),"s",""),"t",""),"u",""),"v",""),"w",""),"x",""),"y",""),"z","")
## 63. Remove numbers from string
To remove numbers from a string (for example Vij1aY A. V4er7ma8 contains numbers
which are not required), we can use nested SUBSTITUTE function to remove numbers. Use
below formula assuming string is in A1 cell -
=SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(
SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(
A1,1,""),2,""),3,""),4,""),5,""),6,""),7,""),8,""),9,""),0,"")
Note - Since this formula is in multiple lines, hence you will have to copy this in Formula
Bar. If you copy this formula in a cell, it will copy this in three rows.
## 64. Roman Representation of Numbers
Use ROMAN function.
## 65. Sum Bottom N Values in a Range
Suppose you have numbers in range A1:A100 and you want to sum up bottom N values
=SUMPRODUCT(SMALL(\$A\$1:\$A\$100,ROW(1:10)))
## In case, you want to ignore 0 values (and blanks)
=SUMPRODUCT(SMALL(IF(\$A\$1:\$A\$100<>0,\$A\$1:\$A\$100),ROW(1:10)))
Both the above formulas will function only if there are at least N values as per ROW(1:N).
Hence, for above formulas, it would work only if there are at least 10 numbers in A1 to
A100.
## Enter the below formulas as Array Formula
=SUM(IFERROR(SMALL(\$A\$1:\$A\$100,ROW(1:10)),0))
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=SUM(IFERROR(SMALL(IF(\$A\$1:\$A\$100<>0,\$A\$1:\$A\$100),ROW(1:10)),0))
Non Array Versions of above formulas (For Excel 2010 and above)
=SUMPRODUCT(AGGREGATE(15,6,\$A\$1:\$A\$100,ROW(1:10)))
=SUMPRODUCT(AGGREGATE(15,6,\$A\$1:\$A\$100/(\$A\$1:\$A\$100<>0),ROW(1:10)))
## 66. Sum Every Nth Row
If your numbers are in range A1:A100, use below formula
=SUMPRODUCT((A1:A100)*(MOD(ROW(A1:A100)-ROW(A1)+1,2)=0))
Above formula is for every 2nd row. Replace 2 with N. Hence, for every 5th row -
=SUMPRODUCT((A1:A100)*(MOD(ROW(A1:A100)-ROW(A1)+1,5)=0))
This is a generic formula and will work for any range. If you range is B7:B50, your formula
would become
=SUMPRODUCT((B7:B50)*(MOD(ROW(B7:B50)-ROW(B7)+1,2)=0))
## 67. We have AVERAGEIF. What about MEDIANIF and
MODEIF?
Excel doesn't provide MEDIANIF and MODEIF. You will have to use Array formulas to
achieve these functionality. Let's assume that our data is like below –
To calculate MEDIANIF and MODEIF, enter below formulas i.e. not by pressing ENTER after
entering your formula but by pressing CTRL+SHIFT+ENTER. This will put { } brackets
around the formula which you can see in Formula Bar. If you edit again, you will have to do
CTRL+SHIFT+ENTER again. Don't put { } manually.
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=MEDIAN(IF(A2:A13="M",B2:B13))
=MODE(IF(A2:A13="M",B2:B13))
Non-Array alternatives
For MEDIANIF
=AGGREGATE(16,6,(B1:B13)/(A1:A13="m"),50%)
For MODEIF
=INDEX(B1:B20,MATCH(MAX(INDEX((COUNTIF(B1:B20,B1:B20)*(A1:A20="m")),,)),IND
EX((COUNTIF(B1:B20,B1:B20)*(A1:A20="m")),,),0))
## 68. Calculate Geometric Mean by Ignoring 0 and Negative
Values
Geometric Mean is a useful mean and is applied only for +ve values. Hence, you will need to
ignore <=0 values while calculating Geometric Mean. It is generally used where %ages are
involved. For example, population growth for first year is 30%, for second year is 25% and
for third year, it is 15%. Then Geometric Mean is used to calculate not Arithmetic Mean.
## Generally, Geometric Mean is calculated by the formula =GEOMEAN(A1:A10)
It would give error if the range contains <=0 values. There are various ways to deal with it
and most commonly used way is to ignore <=0 values while calculating Geometric Mean.
To ignore <=0 values, you must use an Array formula i.e. which must be entered by pressing
CTRL+SHIFT+ENTER.
=GEOMEAN(IF(A1:A10>0,A1:A10))
The above formula takes into account only those values which are positive.
Bonus Tip - When %age growth are involved, you will need to use following ARRAY
formula to calculate Geometric Mean -
=GEOMEAN(IF(A1:A10>0,(1+A1:A10)))-1
## 69. Financial Function - Calculate EMI
You want to take a loan and you want to calculate EMI OR you want to build an EMI
calculator in Excel. It is a fairly easy job to do -
You will need to use PMT function for this. It has following structure -
## PMT(rate, nper, pv, [fv], [type])
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## rate: You rate of interest
nper: No. of payments. Your nper and rate should be on the same scale. i.e if you are
planning to pay up monthly, the rate in your formula should be monthly only. Generally,
interest rate is specified yearly i.e. 10.5% per year. This you should divide by 12 to arrive
at monthly rate. Hence, if you wanted 3 years loan, it means nper would 3x12=36 months.
If it is quarterly, rate = 10.5%/4 = 2.625% and nper would be 3x4 = 12
If it is annual, rate = 10.5% and nper = 3
pv: Your loan amount. You will need to put negative value of this in your formula. If you
don't put negative value, your EMI would be in negative but answer would be same though
with negative sign.
+ve / -ve PMT requires some explanation though you may choose to ignore. It depends upon
your cashflow. If you are taking a loan, hence cash in, hence pv is +ve. But every month, you
will have to pay up something, hence cash out. Hence, PMT is -ve. If you are investing, hence
cash out. Hence pv is -ve. But every month, you will be receiving something, hence cash in.
Hence, PMT is +ve.
Now what is +ve or -ve is simply your preference. I recommend you should not worry about
this.
fv: Your remaining value after you finish your installment. Generally, it is 0 as any lender
will like to recover its money fill. (Default is 0)
type: 0 - At the end of the period, 1 - At the beginning of the period (Default is 0)
Also note, fv and type are optional and may not be required in your formula.
## The formula used in the below picture is =PMT(B1/12,B2,-B3,B4,B5)
Bonus Tip = If you use ABS function, then there would be no need to put negative value of
PV. Hence, formula in this case would be =ABS(PMT(B1/12,B2,B3,B4,B5))
## 70. Financial Function - Calculate Interest Part of an EMI
Now the EMI for a month = Interest for that month and Principal for that month. IPMT is
used to calculate the interest portion of your EMI.
Excel defines IPMT as "Returns the interest payment for a given period for an investment
based on periodic, constant payments and a constant interest rate"
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## The syntax of IPMT is IPMT(rate, per, nper, pv, [fv], [type]).
rate: You rate of interest
## per: Period for which you want to calculate Interest
nper: No. of payments. Your nper and rate should be on the same scale. i.e if you are
planning to pay up monthly, the rate in your formula should be monthly only. Generally,
interest rate is specified yearly i.e. 10.5% per year. This you should divide by 12 to arrive
at monthly rate. Hence, if you wanted 3 years loan, it means nper would 3x12=36 months.
If it is quarterly, rate = 10.5%/4 = 2.625% and nper would be 3x4 = 12
If it is annual, rate = 10.5% and nper = 3
pv: Your loan amount. You will need to put negative value of this in your formula. If you
don't put negative value, your EMI would be in negative but answer would be same though
with negative sign.
+ve / -ve IPMT requires some explanation though you may choose to ignore. It depends
upon your cashflow. If you are taking a loan, hence cash in, hence pv is +ve. But every month,
you will have to pay up something, hence cash out. Hence, IPMT is -ve. If you are investing,
hence cash out. Hence pv is -ve. But every month, you will be receiving something, hence
cash in. Hence, IPMT is +ve.
Now what is +ve or -ve is simply your preference. I recommend you should not worry about
this.
fv: Your remaining value after you finish your installment. Generally, it is 0 as any lender
will like to recover its money fill. (Default is 0)
type: 0 - At the end of the period, 1 - At the beginning of the period (Default is 0)
Also note, fv and type are optional and may not be required in your formula.
## The formula used in the below picture is =IPMT(B1/12,B2,B3,-B4,B5,B6)
Also, since Interest will vary every month, hence it makes sense to calculate it for each
month. Columns H & I carry interest for each month.
Bonus Tip = If you use ABS function, then there would be no need to put negative value of
PV. Hence, formula in this case would be =ABS(IPMT(B1/12,B2,B3,B4,B5,B6))
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## 71. Financial Function - Calculate Principal Part of an EMI
Now the EMI for a month = Interest for that month and Principal for that month. IPMT is
used to calculate the interest portion of your EMI. To calculate the principal part of an EMI,
you will need to use PPMT.
Excel defines PPMT as "Returns the payment on the principal for a given period for an
investment based on periodic, constant payments and a constant interest rate."
## per: Period for which you want to calculate Principal
nper: No. of payments. Your nper and rate should be on the same scale. i.e if you are
planning to pay up monthly, the rate in your formula should be monthly only. Generally,
interest rate is specified yearly i.e. 10.5% per year. This you should divide by 12 to arrive
at monthly rate. Hence, if you wanted 3 years loan, it means nper would 3x12=36 months.
If it is quarterly, rate = 10.5%/4 = 2.625% and nper would be 3x4 = 12
If it is annual, rate = 10.5% and nper = 3
pv: Your loan amount. You will need to put negative value of this in your formula. If you
don't put negative value, your EMI would be in negative but answer would be same though
with negative sign.
+ve / -ve PPMT requires some explanation though you may choose to ignore. It depends
upon your cashflow. If you are taking a loan, hence cash in, hence pv is +ve. But every month,
you will have to pay up something, hence cash out. Hence, PPMT is -ve. If you are investing,
hence cash out. Hence pv is -ve. But every month, you will be receiving something, hence
cash in. Hence, PPMT is +ve.
Now what is +ve or -ve is simply your preference. I recommend you should not worry about
this.
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fv: Your remaining value after you finish your installment. Generally, it is 0 as any lender
will like to recover its money fill. (Default is 0)
type: 0 - At the end of the period, 1 - At the beginning of the period (Default is 0)
Also note, fv and type are optional and may not be required in your formula.
## The formula used in the below picture is =PPMT(B1/12,B2,B3,-B4,B5,B6)
Also, since Principal will vary every month, hence it makes sense to calculate it for each
month. Columns H & I carry Principal for each month.
Bonus Tip = If you use ABS function, then there would be no need to put negative value of
PV. Hence, formula in this case would be =ABS(PPMT(B1/12,B2,B3,B4,B5,B6))
## 72. Financial Function - Calculate Number of EMIs to Pay
Up a Loan
You have taken a loan and you know your EMI capability. So, you want to know how many
months will be taken to pay off a loan completely.
It is fairly easy job to do it in Excel. You will need to use NPER function for this.
Excel defines NPER as "Returns the number of periods for an investment based on periodic,
constant payments and a constant interest rate."
## Syntax of NPER is NPER(rate,pmt,pv,[fv],[type]).
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## rate: You rate of interest
pmt: EMI (Payment per period). You will need to put -ve value of this in your formula.
Your pmt and rate should be on the same scale. i.e if you are planning to pay up monthly,
the rate in your formula should be monthly only. Generally, interest rate is specified yearly
i.e. 10.5% per year. This you should divide by 12 to arrive at monthly rate. Hence, if you
wanted 3 years loan, it means nper would 3x12=36 months.
If it is quarterly, rate = 10.5%/4 = 2.625% and nper would be 3x4 = 12
If it is annual, rate = 10.5% and nper = 3
pv: Your loan amount. You will need to put +ve value of this in your formula.
Note - Either PMT or PV should be -ve. Both can't be +ve and -ve at the same time.
+ve / -ve requires some explanation and this can not be ignored. It depends upon your
cashflow. If you are taking a loan, hence cash in, hence pv is +ve. But every month, you will
have to pay up something, hence cash out. Hence, PMT is -ve. If you are investing, hence
cash out. Hence pv is -ve. But every month, you will be receiving something, hence cash in.
Hence, PMT is +ve.
fv: Your remaining value after you finish your installment. Generally, it is 0 as any lender
will like to recover its money fill. (Default is 0)
type: 0 - At the end of the period, 1 - At the beginning of the period (Default is 0)
Also note, fv and type are optional and may not be required in your formula.
## 73. Financial Function - Calculate Interest Rate
You want to take a loan. You know how much loan to take (pmt), you know how many
months you want to pay up (nper) and you want to know effective rate of interest.
Excel makes it easy to do. RATE function is the answer for this.
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Excel defines RATE as "Returns the interest rate per period of an annuity. RATE is calculated
by iteration and can have zero or more solutions. If the successive results of RATE do not
converge to within 0.0000001 after 20 iterations, RATE returns the #NUM! error value."
Syntax of RATE is RATE(nper, pmt, pv, [fv], [type], [guess]).
## nper: Payment periods. Typically in months.
pmt: EMI (Payment per period). You will need to put -ve value of this in your formula.
Your pmt and rate should be on the same scale. i.e if you are planning to pay up monthly,
the pmt in your formula should be monthly only.
pv: Your loan amount. You will need to put +ve value of this in your formula.
Note - Either PMT or PV should be -ve. Both can't be +ve and -ve at the same time.
+ve / -ve requires some explanation and this can not be ignored. It depends upon your
cashflow. If you are taking a loan, hence cash in, hence pv is +ve. But every month, you will
have to pay up something, hence cash out. Hence, PMT is -ve. If you are investing, hence
cash out. Hence pv is -ve. But every month, you will be receiving something, hence cash in.
Hence, PMT is +ve.
fv: Your remaining value after you finish your installment. Generally, it is 0 as any lender
will like to recover its money fill. (Default is 0)
type: 0 - At the end of the period, 1 - At the beginning of the period (Default is 0)
guess: If you omit guess, it is assumed to be 10 percent. If RATE does not converge, try
different values for guess. RATE usually converges if guess is between 0 and 1. Once again,
note that if PMT is monthly, then Guess should also be monthly. Hence, if you are giving
annual interest rate of 12%, guess should be given as 12%/12 = 1%.
Also note, fv, type and guess are optional and may not be required in your formula.
## The formula used in the below picture is =RATE(B1,-B2,B3,B4,B5,B6/12)
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## 74. Financial Function – Calculate Compounded Interest
As part of our Mathematics courses in our childhood, we had learned about Compounded
Interest. The famous formula which we remember is
Compounded Balance = Principal x (1+rate)^N
## This is a fairly easy job to do in Excel. The formula to be used is FV.
Excel help describes FV as "Returns the future value of an investment based on periodic,
constant payments and a constant interest rate".
## rate: Interest rate on which compounding needs to be done
nper: Total number of periods for which compounding needs to be done. Now rate and nper
should be on the same scale. If interest rate is monthly, then nper should be in months. If
interest rate is quarterly, then nper should be in quarter. If interest rate is annual, then nper
should be in years.
pv: This is the initial principal and it has to be specified in -ve. (Note, I have already
discussed significance of +ve and -ve in many previous tips on Financial Functions.)
## The formula used in below picture for Monthly
=FV(B1/12,B3*12,0,-B2)
## The formula used in below picture for Quarterly
=FV(F1/4,F3*4,0,-F2)
## The formula used in below picture for Yearly
=FV(J1,J3,0,-J2)
Calculator.xlsx
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## 75. Financial Function – Calculate Effective Interest
You are applying for a loan and an interest rate has been quoted. The interest rate which is
quoted is called "Nominal Interest Rate". They will quote Nominal Interest Rate in yearly
terms. Hence, if they quote 12% interest for a loan, this is yearly figure. Now, you generally
pay EMIs every month. They simply say that you need to pay 1% monthly interest which
has been derived by annual interest rate / 12 which 12%/12=1% in this case.
But actually interest rate of 1% (nominal monthly interest rate) is compounded every
month, hence your effective interest rate per year becomes higher. But lending financial
institutions doesn't quote this higher rate as it will make your loan cost look higher.
To calculate Effective Interest Rate, Excel has provided a function called EFFECT.
Excel describes EFFECT - Returns the effective annual interest rate, given the nominal
annual interest rate and the number of compounding periods per year.
## Nominal Rate - Annual Interest Rate
npery - Compounding periods in a Year. For monthly payments, it is 12. For quarterly
payments, it is 4.
In the below picture, the effective interest rate is 12.68% for a monthly payment. This may
be a small difference for a year or two, but if you take mortgage on housing which is say for
20 years, this makes hell of a difference.
## The formula used is =EFFECT(B1,B2)
Now, if you are making an investment and making monthly payments, you will be getting
annual return of 12.68% against 12% if you make yearly payment.
Calculator.xlsx
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## 76. Abbreviate Given Names
If you have names given like -
Smith Johnson
Liz lotte
Christy tiara Lewisk
John
And you need to produce abbreviations or acronyms for them like below in all capitals
Smith Johnson - SJ
Liz lotte - LT
Christy tiara Lewisk - CTL
john - J
Then you can use following formula for the same for upto 3 words in the name -
=UPPER(TRIM(LEFT(A1,1)&MID(A1,FIND(" ",A1&"
")+1,1)&MID(A1,FIND("*",SUBSTITUTE(A1&" "," ","*",2))+1,1)))
## 1. LEFT(A1,1) - Extracts the first letter from the first name
2. MID(A1,FIND(" ",A1&" ")+1,1) -
FIND(" ",A1&" ") - Find finds the first space in the given name to locate the start of the
middle name. " " has been concatenated at the end of A1 so that if there is only first name,
FIND will not give error as it will always find the blanks. +1 has been added to start the MID
position from where the middle name starts.
3. MID(A1,FIND("*",SUBSTITUTE(A1&" "," ","*",2))+1,1))
SUBSTITUTE(A1&" "," ","*",2) will replace the second blank with a *, hence we can find the
position of * to locate the start of last name. As in 2 above, a double space " " has been added
in A1 so that FIND always finds the second space. +1 has been added to start the MID
position from where the last name starts.
4. TRIM will remove all blanks inserted because of 2 or 3.
5. UPPER will convert the string to all capitals.
Note - If you don't to use the concatenation of single space and double space as in 2 and 3,
then IFERROR block can be used. In this case, the formula would become -
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=UPPER(TRIM(LEFT(A1,1)&IFERROR(MID(A1,FIND("
",A1)+1,1),"")&IFERROR(MID(A1,FIND("*",SUBSTITUTE(A1," ","*",2))+1,1),"")))
Note - This technique can be used to extend up to many words. Only change will be in last
block where you can replace 2 with 3, 4,5 and so on in
IFERROR(MID(A1,FIND("*",SUBSTITUTE(A1," ","*",2))+1,1),"") for 4th, 5th, 6th words and
concatenate them....Hence for upto 6 words, the formula would become
=UPPER(TRIM(LEFT(A1,1)&IFERROR(MID(A1,FIND("
",A1)+1,1),"")&IFERROR(MID(A1,FIND("*",SUBSTITUTE(A1," ","*",2))+1,1),"")
&IFERROR(MID(A1,FIND("*",SUBSTITUTE(A1,"
","*",3))+1,1),"")&IFERROR(MID(A1,FIND("*",SUBSTITUTE(A1," ","*",4))+1,1),"")
&IFERROR(MID(A1,FIND("*",SUBSTITUTE(A1," ","*",5))+1,1),"")))
## 77. Get Column Name for a Column Number
Let's suppose, you have a number in A1 and you want to get the column Name for that.
## Hence, if A1=1, you want "A"
Hence, if A1 =26, you want "Z"
Hence, if A1=27, you want "AA" and so on.
## 78. Get Column Range for a Column Number
Let's suppose, you have a number in A1 and you want to get the column range for that.
## Hence, if A1=1, you want "A:A"
Hence, if A1 =26, you want "Z:Z"
Hence, if A1=27, you want "AA:AA" and so on.
## 79. Find the nth Largest Number when there are
duplicates
You know the LARGE function which can find the nth largest value. Hence, if you have a
series like below -
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## Now, if we have a series like below
Now, you give =LARGE(A1:A10,3) and now the result is 24. The reason is that large function
gives the nth largest value in a sorted array. Hence, LARGE function will sort the above array
as {24,24,24,22,22,18,18,9} and 3rd largest is 24.
But actually you want the unique 3rd largest which is 18 as the answer.
## The formula for such case would be
=LARGE(IF(FREQUENCY(\$A\$2:\$A\$10,\$A\$2:\$A\$10)<>0,\$A\$2:\$A\$10),3)
## 80. COUNTIF for non-contiguous range
All of us love COUNTIF. And it is very easy to do - just say =COUNTIF("A1:A100",">5") and
it finds all the values within the range A1 to A100 which are greater than 5. But what if I
wanted the result for only A3, A8 and it should omit other cells. Try putting in following
formula -
## =COUNTIF((A3, A8),">5") and it will give you #VALUE error.
A possible solution is
=(A3>5)+(A8>5)
What happens if you need to do for A3, A4, A5, A8, A24, A40, A45, A89. Now, you will have
to use a formula like -
=(A3>5)+(A4>5)+(A5>5)+(A8>5)+(A24>5)+(A40>5)+(A45>5)+(A89>5)
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The formula becomes cumbersome as the number of cells increase. In this case, you can use
below formula. This single formula can take care of contiguous (like A3:A5) and non-
contiguous ranges both -
=SUM(COUNTIF(INDIRECT({"A3:A5","A8","A24","A40","A45","A89"}),">5"))
## 81. Count the Number of Words in a Cell / Range
Suppose you have been given the following and you need to count the number of words in
a cell or in a range.
## Formula for calculating number of words in a cell -
=LEN(TRIM(A1))-LEN(SUBSTITUTE(TRIM(A1)," ",""))+(TRIM(A1)<>"")
## Formula for calculating number of words in a range -
=SUMPRODUCT(LEN(TRIM(A1:A100))-LEN(SUBSTITUTE(TRIM(A1:A100),"
",""))+(TRIM(A1:A100)<>""))
82. Numerology Sum of the Digits aka Sum the Digits till
the result is a single digit
In Numerology, it is often a task to add the digits till the result is a single digit. For example,
74 = 7 + 4 = 11 = 1 + 1 = 2
23 = 2 + 3 = 5
78 = 7 + 8 = 15 = 1 + 5 = 6
1234567 = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 = 2+ 8 = 10 = 1+ 0 = 1
## The formula to achieve the same is
=MOD(A1-1,9)+1
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## 83. Generate Sequential Numbers and Repeat them
Suppose, you have been given the task to generate a sequence of numbers and repeat them.
For example -
1,2,3,4,1,2,3,4,1,2,3,4
## You can use the below formula and drag down -
=MOD(ROWS(\$1:1)-1,4)+1
Replace 4 with with any other number to generate any other sequence. Hence, if you want
to generate 1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10 then formula
becomes -
=MOD(ROWS(\$1:1)-1,10)+1
## The structure of the formula is
=MOD(ROWS(\$1:1)-1,X)+Y
X - Number of numbers
Y - Starting Number
## Utilizing above formula, you want to generate the sequence
5,6,7,8,9,10,5,6,7,8,9,10,5,6,7,8,9,10, then use below formula (You need 6 numbers and
stating number is 5)
=MOD(ROWS(\$1:1)-1,6)+5
## 84. Repeat a Number and Increment and Repeat....
Suppose, you have been given the task of repeating a number and increment that number
and repeat it. For example -
1,1,1,1,2,2,2,2,3,3,3,3.....(Here, we are repeating it 4 times and incrementing and repeating
4 times again and so on)
## Then you can use following formula
=ROUNDUP(ROWS(\$1:1)/4,0)
Suppose, you want to start the number with 5 not 1, then you can use below formula -
=ROUNDUP(ROWS(\$1:1)/4,0)+4
## Hence, general structure of the formula is
=ROUNDUP(ROWS(\$1:1)/X,0)+Y-1
## X - Number of times a particular number is repeated
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Y - Starting Numbers
Hence, if you want to start with number 7 and you want to repeat it 5 times, then following
formula should be used
=ROUNDUP(ROWS(\$1:1)/5,0)+6
## 85. Generate Non Repeating Random Numbers through
Formula
Suppose, you want to generate non-repeating random numbers between 1 to 30, you can
use following formula in A2 and drag down
=IFERROR(AGGREGATE(14,6,ROW(\$1:\$30)*NOT(COUNTIF(\$A\$1:\$A1, ROW(\$1:\$30))),
RANDBETWEEN(1,30-ROWS(\$1:1)+1)),"")
Note: \$A\$1:\$A1 is with reference to A2 as you put formula in A2 and dragged down.
Suppose, you had put the formula in G4, this should be replaced with \$G\$3:\$G3.
If your starting and ending numbers are in B1 and C1, use below formula
=IFERROR(AGGREGATE(14,6,ROW(INDIRECT(\$B\$1&":"&\$C\$1))*
NOT(COUNTIF(\$A\$1:\$A1,ROW(INDIRECT(\$B\$1&":"&\$C\$1)))),
RANDBETWEEN(\$B\$1,\$C\$1-ROWS(\$1:1)+1)),"")
For versions, prior to 2010 following basic construct can be used (Build error handling
depending upon the version. For example, Excel 2007 will support IFERROR whereas 2003
supports ISERROR) -
=LARGE(INDEX(ROW(\$1:\$30)*NOT(COUNTIF(\$A\$1:\$A1, ROW(\$1:\$30))),,),
RANDBETWEEN(1,30-ROW(A1)+1))
## 86. Extract User Name from an E Mail ID
Assuming A1 has a mail ID say A1:=v.a.verma@gmail.com and you need to retrieve
v.a.verma which is user name in the mail ID. Use following formula –
=IFERROR(LEFT(A1,SEARCH("@",A1)-1),"")
## 87. Extract Domain Name from an E Mail ID
If you want to retrieve domain name which in above example is gmail.com, use following
formula –
=REPLACE(A1,1,SEARCH("@",A1)+1,"")
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## 88. Location of First Number in a String
Suppose you have A1: = “abfg8ty#%473hj” and you want to find what is the position of first
number in this. In this string, first number is 8 and its position is 5. You can use following
formula -
=IFERROR(AGGREGATE(15,6,FIND({1,2,3,4,5,6,7,8,9,0},A1,ROW(INDIRECT("1:"&LEN(A1)
))),1),"")
## 89. Location of Last Number in a String
In the above example, last number is 3 and its position is 12. You can use following formula
to find this –
=IFERROR(AGGREGATE(14,6,FIND({1,2,3,4,5,6,7,8,9,0},A1,ROW(INDIRECT("1:"&LEN(A1)
))),1),"")
## 90. Find the Value of First Non Blank Cell in a Range
=IFERROR(INDEX(A1:A10,MATCH(TRUE,INDEX(NOT(ISBLANK(A1:A10)),,),0)),"")
## 91. Find First Numeric Value in a Range
=IFERROR(INDEX(A1:A100,MATCH(1,INDEX(--ISNUMBER(A1:A100),,),0)),””)
## 92. Find Last Numeric Value in a Range
=IFERROR(1/LOOKUP(2,1/A1:A100),””)
## 93. Find First non Numeric Value in a Range
=IFERROR(INDEX(A1:A100,MATCH(1,INDEX(--ISTEXT(A1:A100),,),0)),””)
## 94. Find Last non Numeric Value in a Range
=IFERROR(LOOKUP(REPT("z",255),A1:A100),””)
## 95. Find Last Used Value in a Range
= IFERROR(LOOKUP(2,1/(A1:A100<>""),A1:A100),””)
96. MAXIF
Note – Excel 2016 has introduced MAXIFS function
Suppose you want to find the Maximum Sales for East Region i.e. MAXIF
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=SUMPRODUCT(MAX((A2:A100="East")*(B2:B100)))
=AGGREGATE(14,6,(\$A\$2:\$A\$100="East")*(\$B\$2:\$B\$100),1)
## SUMPRODUCT formula is faster than the second formula.
97. MINIF
Note – Excel 2016 has introduced MINIFS function
Suppose you want to find the Minimum Sales for West Region i.e. MINIF
=AGGREGATE(15,6,1/(\$A\$2:\$A\$10="West")*(\$B\$2:\$B\$10),1)
But the above formula will not ignore blanks or 0 values in your range. If you want to ignore
0 values /blanks, in your range, then use following formula
=AGGREGATE(15,6,1/((\$A\$2:\$A\$10="West")*(\$B\$2:\$B\$10<>0))*(\$B\$2:\$B\$10),1)
## 98. Generate a Conditional List
Suppose, we have been given a list like below and we want to generate a list based on the
condition Sales > 10
Item Code Sales
A 8
A 16
B 10
B 15
C 5
D 1
D 14
D 12
D 5
## Put below formula and drag down –
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=IFERROR(INDEX(\$A\$2:\$A\$20,AGGREGATE(15,6,(ROW(\$A\$2:\$A\$20)-
ROW(\$A\$2)+1)/(\$B\$2:\$B\$20>10),ROWS(\$1:1))),"")
## 99. Generate a Unique List out of Duplicate Entries
Suppose, you have entries in A2:A100 and you want to generate a list containing only
unique entries in column C starting C2. You can use following formula in C2 and drag down
the formula –
## Case 1 – A2:A100 doesn’t contain any blanks
=IFERROR(INDEX(\$A\$2:\$A\$100,MATCH(0,INDEX(COUNTIF(\$C\$1:\$C1,\$A\$2:\$A\$100),0,0),
)),"")
Case 2 – A2:A100 contains blanks. In this case, you will have to use Array formula.
## =IFERROR(INDEX(\$A\$2:\$A\$100, MATCH(0, IF(\$A\$2:\$A\$100<>"",COUNTIF(C1:\$C\$1,
\$A\$2:\$A\$100)), 0)),"")
OR
=IFERROR(INDEX(\$A\$2:\$A\$100,MATCH(0,COUNTIF(\$C\$1:\$C1,\$A\$2:\$A\$100&""),0)),"")
Note - Array Formula is not entered by pressing ENTER after entering your formula but by
pressing CTRL+SHIFT+ENTER. If you are copying and pasting this formula, take F2 after
pasting and CTRL+SHIFT+ENTER. This will put { } brackets around the formula which you
can see in Formula Bar. If you edit again, you will have to do CTRL+SHIFT+ENTER again.
Don't put { } manually.
## ---- End of Document ----
© eforexcel.com Page 40 of 40 | 19,487 | 72,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-22 | latest | en | 0.350181 |
https://socratic.org/questions/is-5-1-a-solution-to-this-system-of-equations-3y-2x-13-and-4x-8y-12 | 1,579,514,800,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250598217.23/warc/CC-MAIN-20200120081337-20200120105337-00182.warc.gz | 667,072,613 | 6,322 | # Is (-5, -1) a solution to this system of equations 3y = -2x - 13 and -4x = -8y+12?
Feb 25, 2017
Because $\left(- 5 , - 1\right)$ is a solution for both equations it is also a solution for the system of those two equations.
#### Explanation:
First, substitute $\textcolor{red}{- 5} \mathmr{and}$color(blue)(-1)$f \mathmr{and}$color(red)(x) and $\textcolor{b l u e}{y}$ in the first equation and see if both sides of the equation calculate to the same value:
$3 \textcolor{b l u e}{y} = - 2 \textcolor{red}{x} - 13$ becomes:
$3 \times \textcolor{b l u e}{- 1} = \left(- 2 \times \textcolor{red}{- 5}\right) - 13$
$- 3 = 10 - 13$
$- 3 = - 3$
The first, equation has $\left(- 5 , - 1\right)$ as a solution. Now, substitute $\textcolor{red}{- 5} \mathmr{and}$color(blue)(-1)$f \mathmr{and}$color(red)(x) and $\textcolor{b l u e}{y}$ in the second equation and see if both sides of the equation calculate to the same value:
$- 4 \textcolor{red}{x} = - 8 \textcolor{b l u e}{y} + 12$ becomes:
$- 4 \times \textcolor{red}{- 5} = \left(- 8 \times \textcolor{b l u e}{- 1}\right) + 12$
$20 = 8 + 12$
$20 = 20$ | 421 | 1,115 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 16, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2020-05 | latest | en | 0.734202 |
https://earldouglas.com/posts/scala-covariance.html | 1,500,806,826,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424549.12/warc/CC-MAIN-20170723102716-20170723122716-00152.warc.gz | 649,195,704 | 2,186 | # Covariance in Scala
August 14, 2014
## Exercise
Starting with the `Option` implementation from the previous exercise:
``````sealed trait Option[A] { ... }
case class Some[A](a: A) extends Option[A]
case class None[A]() extends Option[A]``````
Make `Option` covariant in its type `A` so that only a single `None` instance is needed:
``case object None extends Option[???]``
``````def quotient(divisor: Int)(dividend: Int): Int =
dividend / divisor
def remainder(divisor: Int)(dividend: Int): Option[Int] =
(dividend % divisor) match {
case 0 => None
case r => Some(r)
}
val x3 =
for {
x <- Some(42)
q = quotient(2)(x)
r <- remainder(4)(x)
} yield (x,q,r)
println(x3) // Some((42,21,2))`````` | 211 | 702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-30 | longest | en | 0.634165 |
https://learn.financestrategists.com/finance-terms/period-expenses/ | 1,669,613,820,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00352.warc.gz | 392,118,754 | 40,324 | # Period Expenses Definition
Period expenses are costs that help a business or other entity generate revenue, but aren’t part of the cost of goods sold.
In general, period expenses include items such as rent, utilities, insurance, and property taxes. They can also include legal fees and loan interest if these amounts are paid in advance.
What a company expects to pay during a particular accounting period is included in an expense account while what it actually pays during the period goes into a prepaid expense account.
## Why Are Period Expenses Important to Know About?
Knowing about how much money a business spends on periods expenses helps its owners and managers understand where their cash flows from operations come from and where they go when operations end up with cash deficits.
The less cash they spend on periods expenses, the more they can save for their own uses or invest in their business operations.
## How Are Period Expenses Calculated?
Period expenses are usually calculated by adding together all expected payments for a period, then subtracting any amounts that were actually paid early. What remains is the total amount of expected expenditures during the period.
For example, Company XYZ’s year-end income statement for its 2016 fiscal year showed \$275,000 in periods expenses.
During the fourth quarter of 2016, Company XYZ expected to pay \$150,000 in rent and utilities and \$100,000 in insurance and property taxes. What is actually paid during that period was \$100,000 in rent and utilities, but only \$10,000 in insurance and property taxes because a storm damaged the roof of one of its properties.
What remains is the total amount of expected expenditures during the period:
(\$150,000 + \$100,000 – \$10,000).
\$160,000
## Period Expense vs Operating Expense
Operating expenses are costs that businesses expect to incur in their attempts to generate revenue. Examples include production materials consumed in making a product and commissions paid to salespeople.
Unlike period expenses, operating expenses often cannot be easily identified by when payments are actually received or made during the accounting periods that they affect.
This means that they can’t be matched with particular sources of cash flows or classified according to whether they involve cash inflows or outflows.
## Period Cost vs Product Expense
Period costs are not the same as product expenses.
Product expenses are part of the cost of producing or acquiring an asset. They are also included in determining the amount of revenue that has been earned when an asset is sold, which in turn can affect both revenues and costs in future accounting periods.
## Types of Period Costs That Should Be Monitored
There are types of period costs that may not be included in the financial statements but are still monitored by the management.
These costs include items that are not related directly to the primary function of a business, such as paying utility bills or filing legal suits.
However, if these costs become excessive they can add significantly to total expenses and they should be monitored closely so managers can take action to reduce them when possible.
Managers will definitely want to find ways to reduce or eliminate these types of costs:
1. Vendors extending lines of credit to businesses which might make postponing or avoiding payment of bills easier
2. The cost of doing business in remote locations because it may be necessary to maintain a presence in an area even if not generating revenue
3. Paying expenses that are the result of decisions made prior to the current period
4. Borrowing money to cover periods costs that do not reduce future expenses or income
## Ways to Reduce or Eliminate These Types of Costs
In some cases, it will be too expensive for a company to eliminate certain types of period costs from its operations completely. In these cases, a more feasible alternative is to try and reduce the amount paid in relation to earlier years.
For example, if your business had been spending \$4000 a year on a particular type of period costs and it’s currently spending \$8000, then you will need to find ways to reduce that expenditure down to where it was previously or at least get the increase in cost under control.
The following are some ways companies can plan upfront to avoid incurring periods expenses as much as possible:
1. Management team diversity – Having a group of decision-makers who come from different backgrounds and with different perspectives will be able to spot potential issues before they become serious problems.
2. Skill development – Having the right skillsets in place will make it easier for decision-makers to notice when things are starting to go wrong before they spiral out of control.
3. Technology – Having cutting-edge technology can be an effective way of avoiding costs like these by allowing better insight into what’s going on at every level within the organization.
4. Inventory and supply chain management – A well-managed inventory will allow the company to reduce the amount of money spent on storage for items that aren’t generating revenue.
5. Ongoing communication with suppliers – By staying in constant communication with suppliers, businesses can make sure they are aware when period costs are starting to get out of hand.
6. Asset management – As the business grows, the right asset management practices can allow spending to be increased in line with increased revenue.
7. Outsourcing non-core activities – If a business is not core to its operations, then outsourcing those responsibilities could help it reduce period expenses.
## Final Thoughts
Period Costs are costs that are incurred during a particular accounting period and may or may not benefit future periods.
These costs should be monitored closely so managers can find ways to reduce the amount paid when possible.
Businesses can plan ahead by diversifying decision-making teams, skill development, using technology effectively, and engaging in ongoing communication with suppliers and other stakeholders.
Managers are always on the lookout for ways to reduce costs while trying to improve the overall effectiveness of their operations. Period expenses are just one category of expense that can have a direct impact on both reducing costs and increasing revenue, so it’s important to keep them in mind when looking for opportunities to improve your business.
Period expenses are costs that help a business or other entity generate revenue, but aren’t part of the cost of goods sold.
Period expenses are important to know about because they can have a direct impact on both reducing costs and increasing revenue. They are also indicators of problems that need to be addressed.
Operating expenses are expenses related to daily operations, whereas period expenses are those costs that have been paid during the current accounting period but will benefit future periods.
Examples of period expenses include vendor bills, storage for supplies or inventory not generating revenue, borrowing money to cover current costs, etc
Management can plan ahead by diversifying decision-making teams, skill development, using technology effectively, and engaging in ongoing communication with suppliers and other stakeholders.
## About the AuthorTrue Tamplin, BSc, CEPF®
True Tamplin is a published author, public speaker, CEO of UpDigital, and founder of Finance Strategists.
True is a Certified Educator in Personal Finance (CEPF®), author of The Handy Financial Ratios Guide, a member of the Society for Advancing Business Editing and Writing, contributes to his financial education site, Finance Strategists, and has spoken to various financial communities such as the CFA Institute, as well as university students like his Alma mater, Biola University, where he received a bachelor of science in business and data analytics.
To learn more about True, visit his personal website, view his author profile on Amazon, or check out his speaker profile on the CFA Institute website. | 1,508 | 8,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-49 | latest | en | 0.973016 |
http://reference.wolfram.com/mathematica/ref/QuartileDeviation.html | 1,398,399,004,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00542-ip-10-147-4-33.ec2.internal.warc.gz | 295,418,189 | 7,232 | BUILT-IN MATHEMATICA SYMBOL
# QuartileDeviation
QuartileDeviation[list]
gives the quartile deviation or semi-interquartile range of the elements in list.
QuartileDeviation[dist]
gives the quartile deviation or semi-interquartile range of the symbolic distribution dist.
## ExamplesExamplesopen allclose all
### Basic Examples (2)Basic Examples (2)
Quartile deviation for a list of exact numbers:
Out[1]=
Quartile deviation of a symbolic distribution:
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http://mac2.microsoft.com/help/office/14/en-us/excel/item/dfecbba6-a4cf-46b2-a4cd-a60f30868cde?category=4cc82111-2d97-4a29-84cb-9f7b75536fa2 | 1,369,451,964,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705352205/warc/CC-MAIN-20130516115552-00008-ip-10-60-113-184.ec2.internal.warc.gz | 159,927,314 | 9,931 | # Chart types
Many chart types are available to help you display data in ways that are meaningful to your audience. Here are some examples of the most common chart types and how they can be used.
## Column chart
Data that is arranged in columns or rows on an Excel sheet can be plotted in a column chart. In column charts, categories are typically organized along the horizontal axisTypically, a line that borders one side of the plot area in a chart, providing a frame of reference for measurement or comparison. For most charts, category labels are plotted along the category axis, which is usually horizontal (x), and data values are plotted along the value axis, which is usually vertical (y). If the data that you want to use for a chart doesn't include data labels, Excel uses numbers (starting at 1) to label the rows or columns for you. Time in days, months, or years is plotted on a time-scale axis. When you rest the pointer over an axis, Excel displays the axis type. and values along the vertical axis.
Column charts are useful to show how data changes over time or to show comparisons among items.
Column charts have the following chart subtypes:
• Clustered column chart Compares values across categories. A clustered column chart displays values in 2-D vertical rectangles. A clustered column in a 3-D chart displays the data by using a 3-D perspective.
• Stacked column chart Shows the relationship of individual items to the whole, comparing the contribution of each value to a total across categories. A stacked column chart displays values in 2-D vertical stacked rectangles. A 3-D stacked column chart displays the data by using a 3-D perspective. A 3-D perspective is not a true 3-D chart because a third value axis (depth axis) is not used.
• 100% stacked column chart Compares the percentage that each value contributes to a total across categories. A 100% stacked column chart displays values in 2-D vertical 100% stacked rectangles. A 3-D 100% stacked column chart displays the data by using a 3-D perspective. A 3-D perspective is not a true 3-D chart because a third value axis (depth axis) is not used.
• 3-D column chart Uses three axes that you can change (a horizontal axis, a vertical axis, and a depth axis). They compare data points along the horizontal and the depth axes.
• Cylinder, cone, and pyramid chart Available in the same clustered, stacked, 100% stacked, and 3-D chart types that are provided for rectangular column charts. They show and compare data in the same manner. The only difference is that these chart types display cylinder, cone, and pyramid shapes instead of rectangles.
## Line chart
Data that is arranged in columns or rows on an Excel sheet can be plotted in a line chart. Line charts can display continuous data over time, set against a common scale, and are therefore ideal to show trends in data at equal intervals. In a line chart, category data is distributed evenly along the horizontal axis, and all value data is distributed evenly along the vertical axis.
Line charts work well if your category labels are text, and represent evenly spaced values such as months, quarters, or fiscal years.
Line charts have the following chart subtypes:
• Line chart with or without markers Shows trends over time or ordered categories, especially when there are many data points and the order in which they are presented is important. If there are many categories or the values are approximate, use a line chart without markers.
• Stacked line chart with or without markers Shows the trend of the contribution of each value over time or ordered categories. If there are many categories or the values are approximate, use a stacked line chart without markers.
• 100% stacked line chart displayed with or without markers Shows the trend of the percentage each value contributes over time or ordered categories. If there are many categories or the values are approximate, use a 100% stacked line chart without markers.
• 3-D line chart Shows each row or column of data as a 3-D ribbon. A 3-D line chart has horizontal, vertical, and depth axes that you can change.
## Pie chart
Data that is arranged in one column or row only on an Excel sheet can be plotted in a pie chart. Pie charts show the size of items in one data seriesA group of related data points plotted in a chart that originate from rows or columns on a single sheet. Each data series in a chart has a unique color or pattern. You can plot one or more data series in a chart. Pie charts have only one data series., proportional to the sum of the items. The data points in a pie chart are displayed as a percentage of the whole pie.
Consider using a pie chart when you have only one data series that you want to plot, none of the values that you want to plot are negative, almost none of the values that you want to plot are zero values, you don't have more than seven categories, and the categories represent parts of the whole pie.
Pie charts have the following chart subtypes:
• Pie chart Displays the contribution of each value to a total in a 2-D or 3-D format. You can pull out slices of a pie chart manually to emphasize the slices.
• Pie of pie or bar of pie chart Displays pie charts with user-defined values that are extracted from the main pie chart and combined into a secondary pie chart or into a stacked bar chart. These chart types are useful when you want to make small slices in the main pie chart easier to distinguish.
• Exploded pie chart Displays the contribution of each value to a total while emphasizing individual values. Exploded pie charts can be displayed in 3-D format. You can change the pie explosion setting for all slices and individual slices. However, you cannot move the slices of an exploded pie manually.
## Bar chart
Data that is arranged in columns or rows on an Excel sheet can be plotted in a bar chart.
Use bar charts to show comparisons among individual items.
Bar charts have the following chart subtypes:
• Clustered bar chart Compares values across categories. In a clustered bar chart, the categories are typically organized along the vertical axis, and the values along the horizontal axis. A clustered bar in 3-D chart displays the horizontal rectangles in 3-D format. It does not display the data on three axes.
• Stacked bar chart Shows the relationship of individual items to the whole. A stacked bar in 3-D chart displays the horizontal rectangles in 3-D format. It does not display the data on three axes.
• 100% stacked bar chart and 100% stacked bar chart in 3-D Compares the percentage that each value contributes to a total across categories. A 100% stacked bar in 3-D chart displays the horizontal rectangles in 3-D format. It does not display the data on three axes.
• Horizontal cylinder, cone, and pyramid chart Available in the same clustered, stacked, and 100% stacked chart types that are provided for rectangular bar charts. They show and compare data the same manner. The only difference is that these chart types display cylinder, cone, and pyramid shapes instead of horizontal rectangles.
## Area chart
Data that is arranged in columns or rows on an Excel sheet can be plotted in an area chart. By displaying the sum of the plotted values, an area chart also shows the relationship of parts to a whole.
Area charts emphasize the magnitude of change over time, and can be used to draw attention to the total value across a trend. For example, data that represents profit over time can be plotted in an area chart to emphasize the total profit.
Area charts have the following chart subtypes:
• Area chart Displays the trend of values over time or other category data. 3-D area charts use three axes (horizontal, vertical, and depth) that you can change. Generally, consider using a line chart instead of a nonstacked area chart because data from one series can be obscured by data from another series.
• Stacked area chart Displays the trend of the contribution of each value over time or other category data. A stacked area chart in 3-D is displayed in the same manner but uses a 3-D perspective. A 3-D perspective is not a true 3-D chart because a third value axis (depth axis) is not used.
• 100% stacked area chart Displays the trend of the percentage that each value contributes over time or other category data. A 100% stacked area chart in 3-D is displayed in the same manner but uses a 3-D perspective. A 3-D perspective is not a true 3-D chart because a third value axis (depth axis) is not used.
## XY (scatter) chart
Data that is arranged in columns and rows on an Excel sheet can be plotted in an xy (scatter) chart. A scatter chart has two value axes. It shows one set of numeric data along the horizontal axis (x-axis) and another along the vertical axis (y-axis). It combines these values into single data points and displays them in irregular intervals, or clusters.
Scatter charts show the relationships among the numeric values in several data series, or plot two groups of numbers as one series of xy coordinates. Scatter charts are typically used for displaying and comparing numeric values, such as scientific, statistical, and engineering data.
Scatter charts have the following chart subtypes:
• Scatter chart with markers only Compares pairs of values. Use a scatter chart with data markers but without lines if you have many data points and connecting lines would make the data more difficult to read. You can also use this chart type when you do not have to show connectivity of the data points.
• Scatter chart with smooth lines and scatter chart with smooth lines and markers Displays a smooth curve that connects the data points. Smooth lines can be displayed with or without markers. Use a smooth line without markers if there are many data points.
• Scatter chart with straight lines and scatter chart with straight lines and markers Displays straight connecting lines between data points. Straight lines can be displayed with or without markers.
## Bubble chart
A bubble chart is a kind of xy (scatter) chart, where the size of the bubble represents the value of a third variable.
Bubble charts have the following chart subtypes:
• Bubble chart or bubble chart with 3-D effect Compares sets of three values instead of two. The third value determines the size of the bubble marker. You can choose to display bubbles in 2-D format or with a 3-D effect.
## Stock chart
Data that is arranged in columns or rows in a specific order on an Excel sheet can be plotted in a stock chart.
As its name implies, a stock chart is most frequently used to show the fluctuation of stock prices. However, this chart may also be used for scientific data. For example, you could use a stock chart to indicate the fluctuation of daily or annual temperatures.
Stock charts have the following chart sub-types:
• High-low-close stock chart Illustrates stock prices. It requires three series of values in the correct order: high, low, and then close.
• Open-high-low-close stock chart Requires four series of values in the correct order: open, high, low, and then close.
• Volume-high-low-close stock chart Requires four series of values in the correct order: volume, high, low, and then close. It measures volume by using two value axes: one for the columns that measure volume, and the other for the stock prices.
• Volume-open-high-low-close stock chart Requires five series of values in the correct order: volume, open, high, low, and then close.
## Surface chart
Data that is arranged in columns or rows on an Excel sheet can be plotted in a surface chart. As in a topographic map, colors and patterns indicate areas that are in the same range of values.
A surface chart is useful when you want to find optimal combinations between two sets of data.
Surface charts have the following chart subtypes:
• 3-D surface chart Shows trends in values across two dimensions in a continuous curve. Color bands in a surface chart do not represent the data series. They represent the difference between the values. This chart shows a 3-D view of the data, which can be imagined as a rubber sheet stretched over a 3-D column chart. It is typically used to show relationships between large amounts of data that may otherwise be difficult to see.
• Wireframe 3-D surface chart Shows only the lines. A wireframe 3-D surface chart is not easy to read, but this chart type is useful for faster plotting of large data sets.
• Contour chart Surface charts viewed from above, similar to 2-D topographic maps. In a contour chart, color bands represent specific ranges of values. The lines in a contour chart connect interpolated points of equal value.
• Wireframe contour chart Surface charts viewed from above. Without color bands on the surface, a wireframe chart shows only the lines. Wireframe contour charts are not easy to read. You may want to use a 3-D surface chart instead.
## Doughnut chart
Like a pie chart, a doughnut chart shows the relationship of parts to a whole. However, it can contain more than one data series. Each ring of the doughnut chart represents a data series.
Doughnut charts have the following chart subtypes:
• Doughnut chart Displays data in rings, where each ring represents a data series. If percentages are displayed in data labels, each ring will total 100%.
• Exploded doughnut chart Displays the contribution of each value to a total while emphasizing individual values. However, they can contain more than one data series.
In a radar chart, each category has its own value axis radiating from the center point. Lines connect all the values in the same series.
Use radar charts to compare the aggregate values of several data series.
Radar charts have the following chart subtypes:
• Radar chart Displays changes in values in relation to a center point.
• Filled radar chart Displays changes in values in relation to a center point, and fills the area covered by a data series with color.
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http://kidsprojects.info/Product-Science/The-Best-Insulation.php | 1,680,073,844,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948951.4/warc/CC-MAIN-20230329054547-20230329084547-00307.warc.gz | 28,763,725 | 5,111 | Home Chemistry Physics Mathematics Bio Chemistry Plant Biology Aerodynamics Zoology Energy Science Product Science Science Experiments Kids Projects Contact Us
# The Best Insulation
The Objective : Would you like to know what the best insulation is for building a house that conserves energy? Well that is what I have been testing for the past months to figure out. I believe that this project can help and prove to builders the best insulation for houses.
Methods/Materials
To begin my experiment I had to buy all the materials that were needed for the project. I bought most items such as wood, insulation, and dry wall at Home Depot and Lowes.
Then I went to a hardware store and bought the right types of nail. To start the boxes we made three boxes out of wood with thin rectangular frames inside.
I then put fiberglass and foamed plastic in two different boxes and left the other empty. I finally put dry wall walls inside of the creating a small house with insulation at the top also.
After a few days of hard work I was ready to test.
Results
After I finally finished testing I got my answer. The winner of the best insulation/energy saver was the fiberglass with a drop of 8.8 degrees Celsius, followed by the foamed plastic with a drop of 10.2 degrees Celsius.
The worst insulation was the spray foam which dropped 10.6 degrees because of an error of the generator possibly making the freezer cold. In last was the empty control with a 15.7 degrees drop.
Conclusions/Discussion
This experiment#s results have shown me and will show others what truly is the best insulation. Builders can now create more energy saving houses when they identify that fiberglass is the best.
To finish off my experiment I will write my written report and my create my poster to state the best insulation.
So, I realize that I was wrong in my hypothesis but, have learned one of the key points to house building and saving energy to keep warm. Fiberglass insulation!
This project was to built three boxes with different insulations and tested each of them in the freezer for an hour to see the greatest and lowest temperature decrease.
Science Fair Project done By Steven L. Delcarson | 446 | 2,189 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-14 | latest | en | 0.956239 |
https://mathandmovement.com/materials/place-value-hop-p1-thousands/ | 1,653,622,064,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662631064.64/warc/CC-MAIN-20220527015812-20220527045812-00495.warc.gz | 439,157,827 | 64,602 | # Place Value Hop P1 (Thousands)
## View the Activities
View all activities for this material or use the filter below to view activities by grade level.
Click on the activity name to view the full activity details.
Hand a student a tens block. Have the student figure out where to put the tens block on the mat. Have the student figure out how the numerals at the...READ MORE
2. ## Multiplying
Have a student create a number on the mat. Then, have them multiply it by 100, 10, .10 or .01. Which way do the digits shift? Have them move the...READ MORE
3. ## Rounding
Have a student create a number on the mat. Then, have them round the numbers to all of the different place values. Use 0 cards for any place after the...READ MORE
4. ## Expanded Form
Have a student create a number and jump it out on the mat. After reading the number, have them write each numeral’s value on an index card and place it...READ MORE
5. ## Clipboard Math
Clip a math worksheet on a clipboard. Have your student figure out the answers to the place value problems by walking or hopping on the mat.
6. ## Adding or Subtracting Numbers
Create notecards, writing + and - with numbers up to the amount of digits you want them to add or subtract (+125, -56, etc.). Mix up the cards and place...READ MORE
7. ## Bean Bag Decides
Have a student create a number and jump it out on the mat. Then, have them toss a bean bag on one of the number cards on the mat. Have...READ MORE
8. ## One Change Leads to Another
Write + 1000, +100, +10, +1, -1000, -100, -10, -1 on notecards. Mix up the cards and add to a large hat or box (or in a pile on the...READ MORE
Hand a student a hundreds block. Have the student figure out where to put the hundreds block on the mat. Have the student figure out how the numerals at the...READ MORE
10. ## The Place Value Walk
Stand on the mat on the ones column. Say, “ones.” Step to the tens, say, “tens.” Then to the hundreds, say, “hundreds,” and to the thousands, say, “thousands.”
Hand a student a ones block. Have the student figure out where to put the ones block on the mat. Have the student figure out how the numerals at the...READ MORE
12. ## Find the Place
Have a student create a number on the mat. Instruct the student to pick up the card in the thousands place, in the tens place, etc. Continue until they have...READ MORE
13. ## Build and Jump Combo
Combine the Building Place Values and Jump out the Number activities. Have students create a number with the number cards. Then, have them build the number using place value blocks...READ MORE
14. ## Jump out the Number
Create a number on the mat, going up to the place value you are working on in class. Have students jump out the number. Example: 5,608 Jump on the 5....READ MORE
15. ## Building Place Values
Once your students are gaining a better understanding of the place values with the activity above, you can have them add on and build the number with blocks or popsicle...READ MORE | 748 | 2,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-21 | latest | en | 0.808658 |
https://id.scribd.com/document/353948790/AEE-03 | 1,563,432,912,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525524.12/warc/CC-MAIN-20190718063305-20190718085305-00208.warc.gz | 414,059,434 | 78,036 | Anda di halaman 1dari 84
Michael E. Auer
Transforms
## Michael E.Auer 21.05.2012 AEE03
AEE Content
Basic Concepts
Three-Phase Circuits
Transforms
Power Conversion and Management
Field Theory
Waves and Vector Fields
Transmission Line Theory
Electrostatics
Magnetostatics
Applications
Magnetic Field Applications
Basics of Electrical Machines
## Michael E.Auer 21.05.2012 AEE03
Chapter Content
Fourier Series
Fourier Transform
Laplace Transform
Applications of Laplace Transform
Z-Transform
## BSC Modul 4: Advanced Circuit Analysis
Fourier Series
Fourier Transform
Laplace Transform
Applications of Laplace Transform
Z-Transform
## Trigonometric Fourier Series (1)
The Fourier series of a periodic function f(t) is a
representation that resolves f(t) into a dc component and
an ac component comprising an infinite series of harmonic
sinusoids.
Given a periodic function f(t) = f(t+nT) where n is an
integer and T is the period of the function.
f (t ) = a0 + (a0 cos n0t + bn sin n0t )
n =1
dc
ac
## and an and bn are as follow
2 2 T
T
an =
T
0
f (t ) cos(no t )dt bn =
T 0
f (t ) sin( no t )dt
## in alternative form of f(t)
f (t ) = a0 + (cn cos(n0t + n )
n =1
dc
ac
bn
where cn = an2 + bn2 , n = tan 1 ( ) (Inverse tangent or arctangent)
an
## Fourier Series Example
Determine the Fourier series of the
waveform shown right. Obtain the
amplitude and phase spectra.
1, 0 < t < 1
f (t ) = and f (t ) = f (t + 2)
0, 1 < t < 2
2 / n , n = odd
2 T An =
T 0
an = f (t ) cos(n0t )dt = 0 and 0, n = even
## 2 / n , n = odd 90, n = odd
2 T
bn = f (t ) sin( n0t )dt = n =
n = even 0, n = even a) Amplitude and
T 0 0, b) Phase spectrum
1 2 1
f (t ) = + sin( nt ), n = 2k 1
2 k =1 n
Truncating the series at N=11
## Three types of symmetry
1. Even Symmetry : a function f(t) if its plot is
f (t ) = f (t )
In this case,
2 T /2
a0 = f (t )dt
T 0
4 T /2
an = f (t ) cos(n0t )dt
T 0
bn = 0
## 2. Odd Symmetry : a function f(t) if its plot is
f (t ) = f (t )
In this case,
a0 = 0
4 T /2
bn =
T
0
f (t ) sin( n0t )dt
## 3. Half-wave Symmetry : a function f(t) if
a0 = 0
4 T /2
T
an = T f (t ) cos(n0t )dt , for n odd
f (t ) = f (t )
0
0 , for an even
2
4 T /2
bn = T
0
f (t ) sin( n0t )dt , for n odd
0 , for an even
## Symmetry Considerations (4)
Example 1
Find the Fourier series expansion of f(t) given
below.
2
1 n n
Ans: f (t ) =
n =1 n
1 cos sin
2 2
t
## Symmetry Considerations (5)
Example 2
Determine the Fourier series for the half-wave
cosine function as shown below.
1 4 1
Ans: f (t ) = 2
2
k =1 n
2
cos nt , n = 2k 1
## Steps for Applying Fourier Series
1. Express the excitation as a Fourier series.
2. Transform the circuit from the time domain to the
frequency domain.
3. Find the response of the dc and ac components in
the Fourier series.
4. Add the individual dc and ac responses using the
superposition principle.
## Circuit Applications (2)
Example
Find the response v0(t) of the circuit below when
the voltage source vs(t) is given by
1 2 1
vs (t ) = + sin (nt ), n = 2k 1
2 n =1 n
## Circuit Applications (3)
Solution
j 2n
Phasor of the circuit V0 =
5 + j 2n
Vs
## For dc component, (n=0 or n=0), Vs = => Vo = 0
2 4 tan 1 2n / 5
For nth harmonic, VS =
n
90, V0 = Vs
25 + 4n 2 2
In time domain,
4 2n
v0 (t ) = cos (nt tan 1 )
k =1 25 + 4n 2 2 5
voltage
## Given: v(t ) = Vdc + Vn cos(n0t Vn ) and i (t ) = Idc + I m cos(m0t Im )
n =1 m =1
1
The average power is P = Vdc I dc + Vn I n cos( n n )
2 n =1
The rms value is Frms = a + (an2 + bn2 )
2
0
n =1
## Average Power and RMS Values (2)
Example
Determine the average power supplied to the
circuit shown below if
i(t)=2+10cos(t+10)+6cos(3t+35) A
## The exponential Fourier series of a periodic function f(t)
describes the spectrum of f(t) in terms of the amplitude and
phase angle of ac components at positive and negative
harmonic.
f (t ) = n
c e
n =
jno t
cn =
1
T
0
T
f (t )e jn0t dt , where 0 = 2 / T
## The plots of magnitude and phase of cn versus n0 are called
the complex amplitude spectrum and complex phase
spectrum of f(t) respectively.
## The complex frequency spectrum of the function
f(t)=et, 0<t<2 with f(t+2)=f(t)
## Application Filter (1)
Filter are an important component of electronics and communications
system.
This filtering process cannot be accomplished without the Fourier series
expansion of the input signal.
For example,
(a) Input and output spectra of a lowpass filter, (b) the lowpass filter passes
only the dc component when c << 0
## Application Filter (2)
(a) Input and output spectra of a bandpass filter, (b) the bandpass filter
passes only the dc component when << 0
## BSC Modul 4: Advanced Circuit Analysis
Fourier Series
Fourier Transform
Laplace Transform
Applications of Laplace Transform
Z-Transform
## It is an integral transformation of f(t) from the time domain
to the frequency domain F()
F() is a complex function; its magnitude is called the
amplitude spectrum, while its phase is called the phase
spectrum.
Given a function f(t), its Fourier transform denoted by F(),
is defined by
F ( ) = f (t )e j t
dt
## Definition of Fourier Transform (2)
Example 1:
Determine the Fourier transform of a single
rectangular pulse of wide and height A, as shown
below.
/2
Solution: F ( ) = Ae jt dt
/ 2
A j t / 2
= e
j / 2
2 A e j / 2 e j / 2
=
2j
= A sin c Amplitude spectrum of the
2 rectangular pulse
## Definition of Fourier Transform (3)
Example 2:
Obtain the Fourier transform of the
switched-on exponential function as
shown.
Solution:
e at
, t>0
f (t ) = e at u (t ) =
0, t<0
Hence,
F ( ) = f (t )e j t
dt = e jat e jt dt
= e ( a + j ) t dt
1
=
a + j
## Properties of Fourier Transform (1)
Linearity:
If F1() and F2() are, respectively, the Fourier
Transforms of f1(t) and f2(t)
F [a1 f1 (t ) + a2 f 2 (t )] = a1 F1 ( ) + a2 F2 ( )
Example:
F [sin(0t )] =
1
2j
[ ( ) ( )]
F e j0t F e j0t = j [ ( + 0 ) ( 0 )]
## Properties of Fourier Transform (2)
Time Scaling:
If F () is the Fourier Transforms of f (t), then
F [ f (at )] = F ( ), a is a constant
1
a a
## If |a|>1, frequency compression, or time expansion
If |a|<1, frequency expansion, or time compression
## Properties of Fourier Transform (3)
Time Shifting:
If F () is the Fourier Transforms of f (t), then
F [ f (t t0 )] = e jt0 F ( )
Example:
j 2
[ ]
F e ( t 2 )u (t 2) =
e
1 + j
## Frequency Shifting (Amplitude Modulation):
If F () is the Fourier Transforms of f (t), then
[
F f (t )e j 0 t
] = F ( )
0
Example:
F [ f (t ) cos(0t )] = F ( 0 ) + F ( + 0 )
1 1
2 2
## Properties of Fourier Transform (5)
Time Differentiation:
If F () is the Fourier Transforms of f (t), then the
Fourier Transform of its derivative is
df
F u (t ) = jF ( s )
dt
Example:
F (
d at
e u (t ) )
1
= a + j
dt
## Properties of Fourier Transform (6)
Time Integration:
If F () is the Fourier Transforms of f (t), then the
Fourier Transform of its integral is
F ( )
F f (t )dt =
t
F (0) ( )
j
Example:
F [u (t )] =
1
+ ( )
j
## Circuit Application (1)
Fourier transforms can be applied to circuits with non-sinusoidal
excitation in exactly the same way as phasor techniques being
applied to circuits with sinusoidal excitations.
Y() = H()X()
## By transforming the functions for the circuit elements into the
frequency domain and take the Fourier transforms of the
excitations, conventional circuit analysis techniques could be
applied to determine unknown response in frequency domain.
Finally, apply the inverse Fourier transform to obtain the
response in the time domain.
## Circuit Application (2)
Example:
Find v0(t) in the circuit shown below for
vi(t)=2e-3tu(t)
Solution:
2
The Fourier transform of the input signal is Vi ( ) =
3 + j
V ( ) 1
The transfer function of the circuit is H ( ) = 0 =
Vi ( ) 1 + j 2
Hence,
1
V0 ( ) =
(3 + j )(0.5 + j )
## BSC Modul 4: Advanced Circuit Analysis
Fourier Series
Fourier Transform
Laplace Transform
Applications of Laplace Transform
Z-Transform
## It is an integral transformation of f(t) from the time
domain to the complex frequency domain F(s)
Given a function f(t), its Laplace transform denoted by
F(s), is defined by
F ( s ) = L[ f (t )]= f (t ) e st dt
## Where the parameter s is a complex number
s = + j , real numbers
## When one says "the Laplace transform" without qualification,
the unilateral or one-sided transform is normally intended.
The Laplace transform can be alternatively defined as the
bilateral Laplace transform or two-sided Laplace transform by
extending the limits of integration to be the entire real axis. If
that is done the common unilateral transform simply becomes
a special case of the bilateral transform.
The bilateral Laplace transform is defined as follows:
F ( s ) = L[ f (t )]= f (t ) e dt
st
## Examples of Laplace Transforms (1)
Determine the Laplace transform of each of the following functions shown:
## a) The Laplace Transform of unit step, u(t) is given by
L[u (t )] = F ( s ) =
1
st
1e dt =
0 s
## Examples of Laplace Transforms (2)
b) The Laplace Transform of exponential function, e-at u(t), a>0 is given by
L[u (t )] = F ( s ) =
1
t st
e e dt =
0 s +
## c) The Laplace Transform of impulse function, (t) is given by
L[u (t )] = F ( s ) = (t )e dt = 1
st
0
## Examples of Laplace Transforms (3)
1 1
F (s) = F ( s) = F ( s) =1
s s +
## Properties of Laplace Transform (1)
Linearity:
If F1(s) and F2(s) are, respectively, the Laplace
Transforms of f1(t) and f2(t)
L[a1 f1 (t ) + a2 f 2 (t )] = a1 F1 ( s ) + a2 F2 ( s )
Example:
1
( )
L[cos(t )u (t )] = L e jt + e jt u (t ) = 2
s
2 s +
2
## Properties of Laplace Transform (2)
Scaling:
If F (s) is the Laplace Transforms of f (t), then
L[ f (at )] = F ( )
1 s
a a
Example:
2
L[sin( 2t )u (t )] =
s 2 + 4 2
## Properties of Laplace Transform (3)
Time Shift:
If F (s) is the Laplace Transforms of f (t), then
L[ f (t a )u (t a )] = e as
F (s)
Example:
L[cos( (t a ))u (t a )] = e as
s
s2 + 2
## Properties of Laplace Transform (4)
Frequency Shift:
If F (s) is the Laplace Transforms of f (t), then
Le [ at
]
f (t )u (t ) = F ( s + a )
Example: [
L e at cos(t )u (t ) =] s+a
( s + a) 2 + 2
## Properties of Laplace Transform (5)
Time Differentiation:
If F (s) is the Laplace Transforms of f (t), then the
Laplace Transform of its derivative is
df
L u (t ) = sF ( s ) f (0 )
dt
Time Integration:
If F (s) is the Laplace Transforms of f (t), then the
Laplace Transform of its integral is
t
1
L f (t )dt = F ( s )
0 s
## Initial and Final Values:
The initial-value and final-value properties allow us
to find f(0) and f() of f(t) directly from its Laplace
transform F(s).
s
s0
+ j
1
f (t ) = st
F ( s ) e ds
2j j F = x 1
## Suppose F(s) has the general form of
N (s) numerator polynomial
F (s) =
D( s) denominator polynomial
## The finding the inverse Laplace transform of F(s)
involves two steps:
1. Decompose F(s) into simple terms using partial
fraction expansion.
2. Find the inverse of each term by matching entries
in Laplace Transform Table.
## The Inverse Laplace Transform (3)
Example
Find the inverse Laplace transform of
3 5 6
F (s) = + 2
s s +1 s + 4
Solution:
3
1 1 5 1 6
f (t ) = L L + L 2
s s +1 s +4
= (3 5e t + 3 sin( 2t )u (t ), t 0
## The Laplace transform is useful in solving linear
integro-differential equations.
Each term in the integro-differential equation is
transformed into s-domain.
Initial conditions are automatically taken into
account.
The resulting algebraic equation in the s-domain can
then be solved easily.
The solution is then converted back to time domain.
## Application to Integro-differential Equations (2)
Example:
Use the Laplace transform to solve the differential
equation
d 2 v(t ) dv(t )
2
+6 + 8v(t ) = 2u (t )
dt dt
## Application to Integro-differential Equations (3)
Solution:
Taking the Laplace transform of each term in the given
differential equation and obtain
2
## Substituting v(0) = 1; v' (0) = 2, we have
2 s 2 + 4s + 2 1 1 1
( s + 6s + 8)V ( s ) = s + 4 + =
2
V (s) = +
4 2
+ 4
s s s s+2 s+4
By the inverse Laplace Transform,
1
v(t ) = (1 + 2e 2t + e 4t )u (t )
4
## BSC Modul 4: Advanced Circuit Analysis
Fourier Series
Fourier Transform
Laplace Transform
Applications of Laplace Transform
Z-Transform
## Steps in Applying the Laplace Transform:
1. Transform the circuit from the time domain to
the s-domain
2. Solve the circuit using nodal analysis, mesh
analysis, source transformation, superposition, or
any circuit analysis technique with which we are
familiar
3. Take the inverse transform of the solution and
thus obtain the solution in the time domain.
## Assume zero initial condition for
the inductor and capacitor,
Resistor : V(s)=RI(s)
Inductor: V(s)=sLI(s)
Capacitor: V(s) = I(s)/sC
The impedance in the s-domain
is defined as Z(s) = V(s)/I(s)
is defined as Y(s) = I(s)/V(s)
## Time-domain and s-domain representations of passive
elements under zero initial conditions.
## Circuit Element Models (3)
Non-zero initial condition for the
inductor and capacitor,
Resistor : V(s)=RI(s)
Inductor: V(s)=sLI(s) + LI(0)
Capacitor: V(s) = I(s)/sC + v(0)/s
## Equivalent Circuits in the s-Domain
Parallel and Series Connection
Z1 ( s )
Z (s) Z1 ( s ) Z2 ( s ) Z (s)
Z2 ( s )
Z1 ( s ) Z 2 ( s )
( s ) Z1 ( s ) + Z 2 ( s )
Z= Z (s) =
Z1 ( s ) + Z 2 ( s )
V1 ( s )
V (s) I1 ( s ) I2 ( s ) I (s)
V2 ( s )
( s ) V1 ( s ) + V2 ( s )
V= ( s ) I1 ( s ) + I 2 ( s )
I=
Z (s)
Thvenin and Norton source models VS ( s ) IS ( s ) Z (s)
VS ( s )
VS ( s ) = Z ( s ) I S ( s ) IS ( s ) =
Z (s)
## Michael E.Auer 21.05.2012 AEE03
Introductory Example
Charging of a capacitor
v
V(0) = 0
## Circuit Element Models Examples (1)
Example 1:
Find v0(t) in the circuit shown below, assuming zero initial
conditions.
## Circuit Element Models Examples (2)
Solution:
Transform the circuit from the time domain to the s-domain:
1
u (t )
s
1H sL = s
1 1 3
F =
3 sC s
## Apply mesh analysis, on solving for V0(s):
3 2 3 4t
V0 ( s ) = v0 (t ) = e sin( 2t ) V, t 0
2 ( s + 4) 2 + ( 2 ) 2 2
Inverse transform
## Circuit Element Models Examples (3)
Example 2:
Determine v0(t) in the circuit shown below, assuming zero
initial conditions.
## Circuit Element Models Examples (4)
Example 3:
Find v0(t) in the circuit shown below. Assume v0(0)=5V .
## Circuit Element Models Examples (5)
Example 4:
The switch shown below has been in position b for a long
time. It is moved to position a at t=0. Determine v(t) for t > 0.
Circuit Analysis
## Circuit analysis is relatively easy to do in the s-domain.
By transforming a complicated set of mathematical
relationships in the time domain into the s-domain where we
convert operators (derivatives and integrals) into simple
multipliers of s and 1/s.
This allow us to use algebra to set up and solve the circuit
equations.
In this case, all the circuit theorems and relationships
developed for dc circuits are perfectly valid in the s-domain.
## Circuit Analysis Example (1)
Example:
Consider the circuit below. Find the
value of the voltage across the
capacitor assuming that the value
of vs(t)=10u(t) V and assume that
at t=0, -1A flows through the
inductor and +5V is across the
capacitor.
## Circuit Analysis Example (2)
Solution:
Transform the circuit from time-domain (a) into s-domain (b) using Laplace
Transform. On rearranging the terms, we have
35 30
V1 =
s +1 s + 2
By taking the inverse transform, we get
v1 (t ) = (35e t 30e 2t )u (t ) V
## Circuit Analysis Example (3)
Example:
The initial energy in the circuit below is zero at t=0. Assume that vs=5u(t) V.
(a) Find V0(s) using the Thevenin theorem. (b) Apply the initial- and final-
value theorem to find v0(0) an v0(). (c) Obtain v0(t).
## Michael E.Auer 21.05.2012 AEE03
Transfer Functions
The transfer function H(s) is the ratio of the output response
Y(s) to the input response X(s), assuming all the initial
conditions are zero.
Y (s)
H (s) =
X (s) h(t) is the impulse response function.
Four types of gain:
1. H(s) = voltage gain = V0(s)/Vi(s)
2. H(s) = Current gain = I0(s)/Ii(s)
3. H(s) = Impedance = V(s)/I(s)
4. H(s) = Admittance = I(s)/V(s)
## Transfer Functions Example (1)
Example:
The output of a linear system is y(t)=10e-tcos4t when the input is x(t)=e-tu(t).
Find the transfer function of the system and its impulse response.
Solution:
Transform y(t) and x(t) into s-domain and apply H(s)=Y(s)/X(s), we get
Y (s) 10( s + 1) 2 4
H (s) = = = 10 40
X ( s ) ( s + 1) 2 + 16 ( s + 1) 2 + 16
Apply inverse transform for H(s), we get
## Transfer Functions Example (2)
Example:
The transfer function of a linear system is
2s
H (s) =
s+6
Find the output y(t) due to the input e-3tu(t) and its impulse
response.
## BSC Modul 4: Advanced Circuit Analysis
Fourier Series
Fourier Transform
Laplace Transform
Applications of Laplace Transform
Z-Transform
## Michael E.Auer 21.05.2012 AEE03
Introduction
In continuous systems Laplace transforms play a unique role. They
allow system and circuit designers to analyze systems and predict
performance, and to think in different terms - like frequency
responses - to help understand linear continuous systems.
## Z-transforms play the role in sampled systems that Laplace
transforms play in continuous systems.
## In continuous systems, inputs and outputs are related by differential
equations and Laplace transform techniques are used to solve
those differential equations.
## In sampled systems, inputs and outputs are related by difference
equations and Z-transform techniques are used to solve those
differential equations.
## For right-sided signals (zero-valued for negative time index) the
Laplace transform is a generalization of the Fourier transform of
a continuous-time signal, and the z-transform is a generalization
of the Fourier transform of a discrete-time signal.
Fourier Transform
generalization
## The Z-transform converts a discrete time-domain signal, which is
a sequence of real or complex numbers, into a complex
frequency-domain representation.
It can be considered as a discrete-time equivalent of the Laplace
transform.
There are numerous sampled systems that look like the one shown below.
discrete-time signals
## Definition of the Z-Transform
Let us assume that we have a sequence, yk.
The subscript "k" indicates a sampled time interval and that yk is the
value of y(t) at the kth sample instant.
## yk could be generated from a sample of a time function.
For example: yk = y(kT), where y(t) is a continuous time function, and T is the
sampling interval.
We will focus on the index variable k, rather than the exact time kT, in all that we
do in the following.
Z [ yk ] = yk z k
k =0
## Michael E.Auer 21.05.2012 AEE03
Z-Transform Example
y k = y0 a k
## We get the Z-Transform for y0 = 1
[ ]
k
a 1 z
Z 1 ak = ak z k = = =
k =0 k =0 z 1
a za
z
## Z-Transform of Unit Impulse and Unit Step
Given the following sampled signal Dk:
Dk is zero for k>0, so all those terms are zero.
Dk is one for k = 0, so that
Z [Dk ] = 1
## Given the following sampled signal uk:
uk is one for all k.
Z [uk ] = 1 + z 1 + z 2 + z 3 ... =
z
z 1
## More Complex Example of Z-Transform
Given the following sampled signal fk:
## f k = f (kT ) = e akT sin(bkT )
Z [ fk ] = fk z k
= e akT sin(bkT ) z k
k =0 k =0
Finally:
1 z z
Z [ fk ] = + c = e aT + jbT
2 j z c z c* where
## Michael E.Auer 21.05.2012 AEE03
Inverse Z-Transform
The inverse z-transform can be obtained using one of two methods:
## a) the inspection method,
b) the partial fraction method.
## In the inspection method each simple term of a polynomial in z,
H(z), is substituted by its time-domain equivalent.
## For the more complicated functions of z, the partial fraction method
is used to describe the polynomial in terms of simpler terms, and
then each simple term is substituted by its time-domain equivalent
term. | 6,285 | 19,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-30 | latest | en | 0.747458 |
https://shop.terrysteachingtidbits.com/collections/3rd-grade-math/products/place-value-chart-interactive-paper-version | 1,643,010,432,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304515.74/warc/CC-MAIN-20220124054039-20220124084039-00424.warc.gz | 569,254,206 | 49,730 | # Interactive Place Value Chart
• \$7.00
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Want to make place value hands-on and interactive? This printable chart will have your students moving and manipulating numbers to help them gain a better understanding of their place value all year long! There are printable student versions for individual use.
The large chart is easy to customize and the individual charts come in multiple variations, so you have options! Students can use the large chart on a wall, bulletin board, or white board in conjunction with individual student charts for those at their seats (great inside of a dry erase pocket).
You’ll love how adaptable this resource is! The printable wall chart can be customized to any length to fit the place values or skills that you’re teaching (whole numbers or decimals).
Use this resource during an observation (others have!) to show how your students are able to go beyond worksheets and get hands-on and interactive with place value.
How Can You Use This?
Introduce basic place value skills by reading numbers to your students and having them form the number in standard form on the chart.
As you progress to more complex skills, use a double chart to show how to compare numbers by finding the greatest place value where the digits in the two numbers differ.
In upper elementary, this chart is perfect for showing students how to multiply and divide numbers by powers of 10, as they can shift the digits on the chart to show how their value increases or decreases.
WHAT OTHER TEACHERS ARE SAYING…
⭐️⭐️⭐️⭐️⭐️ This is an amazing resource! After using it just twice, it has truly helped my students to visualize place value, comparing, ordering, and rounding! I cannot wait to continue to use it throughout the year! - Oh Happy Wray
⭐️⭐️⭐️⭐️⭐️ This is one of my all-time favorite interactive bulletin boards. I laminated the place value chart to be able to write with an expo marker. We use it daily and the students love it as much as I do. - Christel P.
⭐️⭐️⭐️⭐️⭐️ Loved using this in my classroom. It is a beautiful interactive display, and the individual charts keep the whole class engaged. It makes place value easy to visualize and understand. - Amy Jo V.
## What You Get
• Place value places from hundred billions down to thousandths (I use two sets to compare two different numbers)
• Number A and Number B signs to allow for digit value comparisons (5.NBT.1)
• Numbers to be used on the board (make as many copies of each number as you need)
• Arrows with x10 and ÷10 (as well as x1/10) to show the increase and decrease in powers of ten as you move across the chart
• Place Value Title
• Individual Student Place Value charts so that they can work at their own desks while another student is using the large chart.
• Templates for students to complete the 4 operations (perfect to laminate or stick inside of dry erase pockets)
• addition - whole numbers and decimals
• subtraction - whole numbers and decimals
• multiplication - 4x1, 4x2, 4x3, 3x3, 3x2, 3x1, 2x2
• division, 2x1, 3x1, 4x1, 2x2, 3x2, 4x2 (with and without remainders)
There are 18 sets (ones to thousands through hundred billions, and thousandths through hundred billions). EACH set includes:
• a single page with one place value chart with and without labels
• a page that includes 3 place value charts (you cut these into strips and can get 3 students' charts from each page - save paper!) with and without labels
• a page with two sets of place value charts for comparison of numbers and digits with and without labels
• a page with two sets of place value charts with a comparison statement (>, <, or =) on the bottom with and without labels)
You will also receive a simpler place value chart that will show each place's name along with its value in both whole number, decimal, and fraction form.
## Customer Reviews
Based on 2 reviews | 897 | 3,946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-05 | latest | en | 0.898587 |
https://www.cfd-online.com/Forums/fluent/114329-what-meant-operating-pressure-fluent.html | 1,716,264,441,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058383.61/warc/CC-MAIN-20240521025434-20240521055434-00325.warc.gz | 610,129,634 | 21,440 | # what is meant by Operating Pressure in Fluent?
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March 9, 2013, 03:51 what is meant by Operating Pressure in Fluent? #1 New Member Mifrah Ali Join Date: Mar 2013 Posts: 3 Rep Power: 13 hello all, i am a new user in Fluent. i want to know what exactly is meant by operating pressure in fluent? its default value is 101325 pa, is it atmospheric pressure? but sometime we also assume 0 as operating pressure, its making me feel dizzy.... what will be the operating pressure in 10 meter depth of a pool?
March 9, 2013, 05:40 #2 Senior Member Join Date: Aug 2011 Posts: 421 Blog Entries: 1 Rep Power: 21 It depends on how you model the density. It is always valid to set the operating pressure to (approximately) the mean pressure of the flow field. But it is only when you model the density using incompressible ideal gas law, or you are dealing with low-Mach-number compressible flow that the value of operating pressure is important. The operating pressure is not used if the density is modeled to be constant. You should check the Fluent user manual because it has more detailed and accurate description of the operating pressure.
March 9, 2013, 12:38 #3 New Member Mifrah Ali Join Date: Mar 2013 Posts: 3 Rep Power: 13 in my case density is constant so it means i can take operating pressure as zero??????
March 9, 2013, 15:16 dear .... #4 Member farzadpourfattah Join Date: Mar 2013 Posts: 41 Rep Power: 13 when you set pressure of operation condition 0 pascal, you prevent reverse flow in your domain.
March 10, 2013, 01:15 #5 Senior Member Join Date: Aug 2011 Posts: 421 Blog Entries: 1 Rep Power: 21 Yes. You can set operating pressure to any value because it is not used in case of constant density. chaitanyaarige likes this.
March 12, 2013, 14:02 #6 New Member Mifrah Ali Join Date: Mar 2013 Posts: 3 Rep Power: 13 Thank you
June 8, 2013, 12:13 #7 Senior Member FHydro Join Date: Jan 2013 Posts: 197 Rep Power: 13 what is that value in VOF model? (water+air)
June 15, 2013, 11:27 vof #8 Member farzadpourfattah Join Date: Mar 2013 Posts: 41 Rep Power: 13 if you set value of water 1 in a zone,value of air will be 0 at same zone.
June 6, 2016, 12:57
#9
New Member
rizky djon hansemit
Join Date: Mar 2016
Posts: 2
Rep Power: 0
Quote:
Originally Posted by blackmask It depends on how you model the density. It is always valid to set the operating pressure to (approximately) the mean pressure of the flow field. But it is only when you model the density using incompressible ideal gas law, or you are dealing with low-Mach-number compressible flow that the value of operating pressure is important. The operating pressure is not used if the density is modeled to be constant. You should check the Fluent user manual because it has more detailed and accurate description of the operating pressure.
can you tell us how do we know the mean pressure of the flow field in fluent. in my case i simulate 2d vawt in a wind tunnel. is there any consideration when we want to specify certain value of operating condition in fluent for incompressible ideal gas air. what about reference pressure location? its default value is [0,0,0]. how to specify the right reference pressure location in our simulation? please help us
June 12, 2016, 17:49 #10 New Member rizky djon hansemit Join Date: Mar 2016 Posts: 2 Rep Power: 0 is there any help?
September 20, 2017, 00:07
#11
New Member
Parag Mangave
Join Date: Sep 2017
Posts: 6
Rep Power: 8
Quote:
Originally Posted by blackmask It depends on how you model the density. It is always valid to set the operating pressure to (approximately) the mean pressure of the flow field. But it is only when you model the density using incompressible ideal gas law, or you are dealing with low-Mach-number compressible flow that the value of operating pressure is important. The operating pressure is not used if the density is modeled to be constant. You should check the Fluent user manual because it has more detailed and accurate description of the operating pressure.
i am trying to simulate flow over blunt bodies my conditions are 100500pa, 303k,1.1551 density, what should be my operating pressure and guage pressure?
September 20, 2017, 09:01 #12 Senior Member Join Date: Nov 2013 Posts: 1,965 Rep Power: 26 You choose! Real pressure = operating pressure + guage pressure. If you want the real pressure to be 100500 Pa, you can set operating pressure at 100500 Pa, and then at the boundary set 0 Pa. Or set the operating pressure at 100000 Pa, and the pressure at the boundary at 500 Pa. Or the operating pressure at 0 Pa, and the pressure at the boundary at 100500 Pa. Physically, it is all the same. Exactly the same, identical. Only numerically, it is helpful to have smaller numbers for guage pressures, so put your operating pressure close to the 'average' pressure in your system. Extreme example: don't set your operating pressure at 10000100500 Pa, and your guage pressure at -10000000000 Pa, you will get numerical problems. tariq, chaitanyaarige, paragmangave and 2 others like this.
September 20, 2017, 09:11 #13 New Member Parag Mangave Join Date: Sep 2017 Posts: 6 Rep Power: 8 Also is prism layer generation necessary for supersonic speed?. Without using prism layer mesh can I use k omega or k epsilon turbulence model?
October 26, 2017, 05:47 #14 New Member Rouven Join Date: Oct 2017 Posts: 1 Rep Power: 0 A Question regarding this Operation Pressure: I have a low-Mach-No. Simulation and i know that the pressure inside my geometry should be s.th. about 7-8bar. So i set the operation Pressure to 7 bar. As result I got a pressure value of 1bar in the interior. So does this mean that I have to add 1 bar to the operation pressure to get the "real" value of pressure? Actually I have 1 bar beyond the inlets and outlets. Does this operation pressure influence my "outside"-conditions? So if I want 1 bar outside, do i have to set the pressure boundary condition to -6bar because 7bar minus 6 bar is 1bar? Best regards
March 9, 2018, 06:11 #15 New Member Hongyang Chen Join Date: Mar 2018 Location: Shanghai Posts: 1 Rep Power: 0 “ANSYS Fluent provides a “floating operating pressure” option to handle time-dependent compressible flows with a gradual increase in the absolute pressure in the domain. This option is desirable for slow subsonic flows with static pressure build-up, since it efficiently accounts for the slow changing of absolute pressure without using acoustic waves as the transport mechanism for the pressure build-up. Examples of typical applications include the following: 1.combustion or heating of a gas in a closed domain 2.pumping of a gas into a closed domain Limitations: The floating operating pressure option should not be used for transonic or incompressible flows. In addition, it cannot be used if your model includes any pressure inlet, pressure outlet, exhaust fan, inlet vent, intake fan, outlet vent, or pressure far field boundaries.“ From ANSYS FLUENT help document.
May 11, 2018, 01:26 #16 New Member BHOLU KUMAR Join Date: May 2018 Posts: 10 Rep Power: 8 Suppose I have set intet pressure = 405300 Pa (Fluent solver) Outlet gauge pressure = 4000 Pa Operating pressure = 0 Pa Then A/C you Real Pressure at inlet = 405300+0 = 405300 Pa outlet pressure gauge pressure = 4000 + 0 = 4000 Pa only ?
May 11, 2018, 05:31
#17
Senior Member
Kushal Puri
Join Date: Nov 2013
Posts: 182
Rep Power: 12
Quote:
Originally Posted by BHOLU Suppose I have set intet pressure = 405300 Pa (Fluent solver) Outlet gauge pressure = 4000 Pa Operating pressure = 0 Pa Then A/C you Real Pressure at inlet = 405300+0 = 405300 Pa outlet pressure gauge pressure = 4000 + 0 = 4000 Pa only ?
Yes if you are using 0 operating pressure the you have to apply absolute value of pressure at outlet also
May 11, 2018, 06:16 #18 New Member BHOLU KUMAR Join Date: May 2018 Posts: 10 Rep Power: 8 Thank you Kushal.
May 16, 2018, 09:14 Operating pressure #19 New Member Nitesh Kumar Join Date: May 2016 Posts: 17 Rep Power: 9 Hi Bholu, Did you get the desired result.
December 4, 2018, 05:59 #20 New Member Fab Join Date: Dec 2018 Posts: 13 Rep Power: 7 Hi all, Thanks for all your answer. What about if i want to calculate in a bend pipe the pressure losses and the poutlet is 10bar in a incompressible fluid? From what i understood: a) operating pressure= 10 bar b) gauge pressure = 0 bar? Correct? Why my pressure in CFD post is not considering the operating pressure? Thanks
Tags fluent, operating pressure | 2,254 | 8,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.9289 |
https://www.traditionaloven.com/tutorials/distance/convert-china-cun-unit-to-ell-english-unit-measure.html | 1,627,433,892,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153515.0/warc/CC-MAIN-20210727233849-20210728023849-00097.warc.gz | 1,089,409,640 | 17,444 | Convert 市寸 to English | Chinese cùn to ells
# length units conversion
## Amount: 1 Chinese cùn (市寸) of length Equals: 0.029 ells (English) in length
Converting Chinese cùn to ells value in the length units scale.
TOGGLE : from ells into Chinese cùn in the other way around.
## length from Chinese cùn to ell conversion results
### Enter a new Chinese cùn number to convert
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many digits after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other length measuring units - complete list.
How many ells are in 1 Chinese cùn? The answer is: 1 市寸 equals 0.029 English
## 0.029 English is converted to 1 of what?
The ells unit number 0.029 English converts to 1 市寸, one Chinese cùn. It is the EQUAL length value of 1 Chinese cùn but in the ells length unit alternative.
市寸/English length conversion result From Symbol Equals Result Symbol 1 市寸 = 0.029 English
## Conversion chart - Chinese cùn to ells
1 Chinese cùn to ells = 0.029 English
2 Chinese cùn to ells = 0.058 English
3 Chinese cùn to ells = 0.087 English
4 Chinese cùn to ells = 0.12 English
5 Chinese cùn to ells = 0.15 English
6 Chinese cùn to ells = 0.17 English
7 Chinese cùn to ells = 0.20 English
8 Chinese cùn to ells = 0.23 English
9 Chinese cùn to ells = 0.26 English
10 Chinese cùn to ells = 0.29 English
11 Chinese cùn to ells = 0.32 English
12 Chinese cùn to ells = 0.35 English
13 Chinese cùn to ells = 0.38 English
14 Chinese cùn to ells = 0.41 English
15 Chinese cùn to ells = 0.44 English
Convert length of Chinese cùn (市寸) and ells (English) units in reverse from ells into Chinese cùn.
## Length, Distance, Height & Depth units
Distance in the metric sense is a measure between any two A to Z points. Applies to physical lengths, depths, heights or simply farness. Tool with multiple distance, depth and length measurement units.
# Converter type: length units
First unit: Chinese cùn (市寸) is used for measuring length.
Second: ell (English) is unit of length.
QUESTION:
15 市寸 = ? English
15 市寸 = 0.44 English
Abbreviation, or prefix, for Chinese cùn is:
Abbreviation for ell is:
English
## Other applications for this length calculator ...
With the above mentioned two-units calculating service it provides, this length converter proved to be useful also as a teaching tool:
1. in practicing Chinese cùn and ells ( 市寸 vs. English ) measures exchange.
2. for conversion factors between unit pairs.
3. work with length's values and properties.
To link to this length Chinese cùn to ells online converter simply cut and paste the following.
The link to this tool will appear as: length from Chinese cùn (市寸) to ells (English) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 815 | 2,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-31 | latest | en | 0.731431 |
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#### Noutie
• Posts: 10
##### Frequency detection
« on: 16 May '06 - 09:48 »
Hi everyone,
I have a quite physical question which came out of an investigation in which I'm currently involved and I hope that somebody can provide me some informationon this topic.
I analyze, by a self written program using BASS, sounds recorded by a microphone. On the received data I apply a Fast Fourier Transformation and I analyse which frequency was played by finding the highest amplitude and it's belonging frequency. I'm capable of correctly reckognizing the tones of a simple song, played by an 'normal' external speaker. Things become quite different when I, for example, want to analyse a mobile phone sound.
When the mobile phone plays an A4 note (440 Hz) things starts well, but after a second or so, not the A4, but the A5 (880 Hz) gets the highest amplitude. Offcourse, a mobile phone speakers is encapsulated, so perhaps this happens due to interference? Or due to the limited frequency response of a mobile phone speaker (300 Hz - 4 kHz)? (Remember: The same soundfile on a 'normal' speaker is recorded like I supposed it would). Clues would be appreciated...
Does anyone has an idea on how to get first harmonic (so the actual tone which is played)? Or ideas on which possibilities are there to apply the highest amplitude to the first harmonic are also more than welcome...
Cheers,
Arnoud
#### Rah'Dick
• XMPlay Support
• Posts: 963
##### Re: Frequency detection
« Reply #1 on: 16 May '06 - 11:59 »
I think the actual problem is that mobile phone speakers are so crappy and have lots of resonance frequencies that aren't there on bigger speakers. To get the real frequency, you need to analyse all peaks that are over some threshold, in the whole frequency range. Try recording the sound you're analysing and have a look at the spectrum (with Adobe Audition for example, or XMPlay's Spectrograph). Sometimes the resonance frequency is just that - a resonance that is nothing like the original tone.
I hope that helps a bit...
#### Torkell
• Posts: 1169
##### Re: Frequency detection
« Reply #2 on: 16 May '06 - 12:08 »
A thought: you could find the highest amplitude, then search for the smallest power-of-two factor of it (so, e.g. if you find 880Hz you then look for 440Hz, 220Hz and 110Hz) that still has a large enough amplitude.
#### Noutie
• Posts: 10
##### Re: Frequency detection
« Reply #3 on: 16 May '06 - 13:02 »
Tnx to you guys for responding so fast!
A thought: you could find the highest amplitude, then search for the smallest power-of-two factor of it (so, e.g. if you find 880Hz you then look for 440Hz, 220Hz and 110Hz) that still has a large enough amplitude.
Believe it or not, but this option just went also through my head...
The smallest power of two factor is easy to search for but the problem is how do I define the large enough amplitude peak? If I cummulate all the amplitudes of the FFT (I use BASS_DATA_FFT4096) and divide it by 4096 to get the average, would this be a nice way to set the large enough amplitude? The lowest frequency with an amplitude peak, is always the first harmonic? I mean when a MIDI tone of 440 hz is played, there will never ever be 220 hz in the spectrum, or isn't this correct?
... Try recording the sound you're analysing and have a look at the spectrum (with Adobe Audition for example, or XMPlay's Spectrograph).
I have done this, and that's why I know the problem resides not in my program... The spectrograph's shows that in the first period of time the frequency with the highest amplitude is the note which is defined in the MIDI file, later on, this frequency amplitude attenuates a little, and the second harmonic amplitude gains above the frequency supposed to have the highest amplitude.
What I also think is strange, (but a bit offtopic) is the fact that I don't hear the A5 (880 Hz) note when the sound is played. Not cocky or something, but my 'audiosensiblity' is above average, but still I'm not capable of hearing the difference between an A4 played in a pure way (only 880 Hz without harmonics), and an A4 played with an overload of harmonics, on which even the A5 is played louder then the A4.
« Last Edit: 16 May '06 - 13:06 by Noutie »
#### Dotpitch
• Posts: 2878
##### Re: Frequency detection
« Reply #4 on: 16 May '06 - 17:44 »
I mean when a MIDI tone of 440 hz is played, there will never ever be 220 hz in the spectrum, or isn't this correct?
If you're have only one source of sound (a speaker), then there can only be harmonics in higher frequencies (since 220 Hz isn't a harmonic of 440 Hz, you can draw out the waves and see that there is no integer number of 220 Hz-waves in one 440 Hz-wave).
It's odd that the first harmonic has a higher amplitude than the ground tone... something is resonating, either in the mobile phone, or in the microphone.
What I also think is strange, (but a bit offtopic) is the fact that I don't hear the A5 (880 Hz) note when the sound is played. Not cocky or something, but my 'audiosensiblity' is above average, but still I'm not capable of hearing the difference between an A4 played in a pure way (only 880 Hz without harmonics), and an A4 played with an overload of harmonics, on which even the A5 is played louder then the A4.
Just try with Audacity (audacity.sourceforge.net). It has a tone generator, and I can clearly hear the difference between 440 Hz, 880 Hz, 440 with 880 Hz and 440 with 880 and 1760 Hz on a normal speaker (though the sound gets annoying ).
#### Rah'Dick
• XMPlay Support
• Posts: 963
##### Re: Frequency detection
« Reply #5 on: 16 May '06 - 23:38 »
Another thought:
What if you're FFT'ing both channels? To get one single-channel frequency, you'd have to divide that by 2, right? | 1,442 | 5,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-47 | latest | en | 0.945097 |
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Автор Bobek, история, 6 месяцев назад, ,
I read some articles about ternary search and it seems that it's very similar to binary search, but we divide interval on three rarther than two parts. Is there any problem that can be solved using ternary search and can't be solved using (or at least much harder) binary search ? When should I use ternary search over binary search ?
•
• -4
•
» 6 месяцев назад, # | ← Rev. 2 → +1 Ternary search is used when you need to find the extreme value of a unimodal function, for example -x^2+3, which has is strictly increasing, then it has a maximum at x=0, then it is strictly decreasing. You can read about this application here.
• » » 6 месяцев назад, # ^ | 0 I read that, but it doesn't help. In binary search we also look for minimum/maximum, so I guess it also applies to unimodal functions
• » » » 6 месяцев назад, # ^ | +2 It can't, because in a binary search needs a function that is monotonic. For example i'll use -x^2 +3. Let's say that binary search has found for some x that value is 2. You can't determine whether you should look for x that is larger or smaller than your current x because x can be both 1 and -1, and the maximum is achieved when x is 0.
• » » » » 6 месяцев назад, # ^ | ← Rev. 3 → +1 Actually you can also use binary search for an unimodal function.For some positive epsilon ε, we look at the values f(x) and f(x + ε). If f(x) < f(x + ε), then x is left of the apex (or close enough to the apex). Otherwise, x is right of the apex. However this can't always be used due to precision issues. | 481 | 1,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-09 | longest | en | 0.821054 |
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Question
# If α and β are roots of x2 - (k+1) x + 12 (k2+k+1) = 0, then α2+β2 is equal
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Solution
## The correct option is B kα and β are roots of the equation x2−(k+1)x+12(k2+k+1)=0.α+β=−−(k+1)1=k+1 ------ ( 1 )αβ=k2+k+12 ----- ( 2 )Now,⇒ (α+β)2=α2+β2+2αβ⇒ (k+1)2=α2+β2+2×k2+k+12⇒ k2+2k+1=α2+β2+k2+k+1⇒ k2+2k+1−k2−k−1=α2+β2∴ α2+β2=k
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Prev Next | 1,272 | 5,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-38 | latest | en | 0.90863 |
http://www.conversion-website.com/mass/quarter-US-to-kilotonne.html | 1,511,368,757,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806615.74/warc/CC-MAIN-20171122160645-20171122180645-00479.warc.gz | 366,236,012 | 4,557 | # Quarters (US) to kilotonnes (qtr to kt)
## Convert quarters (US) to kilotonnes
Quarters (US) to kilotonnes conversion calculator shown above calculates how many kilotonnes are in 'X' quarters (US) (where 'X' is the number of quarters (US) to convert to kilotonnes). In order to convert a value from quarters (US) to kilotonnes (from qtr to kt) type the number of qtr to be converted to kt and then click on the 'convert' button.
## Quarters (US) to kilotonnes conversion factor
1 quarter (US) is equal to 1.133980925E-5 kilotonnes
## Quarters (US) to kilotonnes conversion formula
Mass(kt) = Mass (qtr) × 1.133980925E-5
Example: Pressume there is a value of mass equal to 401 quarters (US). How to convert them in kilotonnes?
Mass(kt) = 401 ( qtr ) × 1.133980925E-5 ( kt / qtr )
Mass(kt) = 0.00454726350925 kt or
401 qtr = 0.00454726350925 kt
401 quarters (US) equals 0.00454726350925 kilotonnes
## Quarters (US) to kilotonnes conversion table
quarters (US) (qtr)kilotonnes (kt)
220.0002494758035
240.000272155422
260.0002948350405
280.000317514659
300.0003401942775
320.000362873896
340.0003855535145
360.000408233133
380.0004309127515
400.00045359237
420.0004762719885
440.000498951607
460.0005216312255
480.000544310844
500.0005669904625
520.000589670081
540.0006123496995
560.000635029318
580.0006577089365
600.000680388555
quarters (US) (qtr)kilotonnes (kt)
1500.0017009713875
2000.00226796185
2500.0028349523125
3000.003401942775
3500.0039689332375
4000.0045359237
4500.0051029141625
5000.005669904625
5500.0062368950875
6000.00680388555
6500.0073708760125
7000.007937866475
7500.0085048569375
8000.0090718474
8500.0096388378625
9000.010205828325
9500.0107728187875
10000.01133980925
10500.0119067997125
11000.012473790175
Versions of the quarters (US) to kilotonnes conversion table. To create a quarters (US) to kilotonnes conversion table for different values, click on the "Create a customized mass conversion table" button.
## Related mass conversions
Back to quarters (US) to kilotonnes conversion
TableFormulaFactorConverterTop | 718 | 2,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2017-47 | latest | en | 0.404712 |
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#### Problem 290. Make one big string out of two smaller strings
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#### Problem 2. Make the vector [1 2 3 4 5 6 7 8 9 10]
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Tags basic, basics, colon
#### Problem 1. Times 2 - START HERE
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Tags intro, math, easy
51 – 71 of 71 | 480 | 1,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-04 | latest | en | 0.764698 |
https://www.plati.ru/itm/7-4-7-the-solution-of-the-problem-of-the-collection-of/2057871?lang=en-US | 1,503,353,399,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109670.98/warc/CC-MAIN-20170821211752-20170821231752-00400.warc.gz | 968,388,052 | 12,121 | # 7.4.7 The solution of the problem of the collection of
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# Description
The position of the crank OA determined by the angle φ = 2t. Determine the acceleration projection ah point A at time t = 1, if the length of OA = 1 m. | 144 | 472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-34 | latest | en | 0.724773 |
https://www.matthewpais.com/research/level-set | 1,726,699,568,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00643.warc.gz | 808,568,345 | 16,462 | # Level Set Method
Introduction
Within the XFEM framework, the discontinuities are modeled independent of the finite element mesh. This creates the problem of how one goes about keeping track of the evolution of these discontinuities as they are not explicitly defined by the finite element mesh. The level set method has been used to tracking cracks, voids and holes. The level set method was originally introduced by Osher and Sethian1 for tracking the evolution of moving boundaries and later adopted and modified for use in XFEM. The narrow band level set method introduced by Adalsteinsson and Sethian2 can be used to reduce the computational burden associated with the level set method. The level set function for a closed curve takes the following form.
Cracks
The level set method was originally introduced for tracking the evolution of closed boundaries. The method was then extended for tracking the evolution of open segments by Stolarska3 so that the evolution of a crack could be tracked. This is accomplished through the use of two orthogonal level set functions. The psi level set is used to track the crack body, while the phi level set is used to track the crack tip. For the case when there are multiple crack tips, multiple phi level set functions are used. The ith crack tip is found to be the intersection of the zero level set of psi and the zero level set of the ith phi. Specifically, the phi and psi level set functions are defined such in the following way.
As the crack grows, the phi and psi level set functions evolve as shown in the animation above. Both the full and narrow band level set representations of the crack are given. If the animation does not play you may need to install Adobe Flash to your computer.
Inclusions and Voids
The level set method can also be used to track the evolution of voids and inclusions4 using the original version of the level set method for closed boundaries. In particular for a circular void or inclusion the cooresponding level set function is given by
where (xo,yo) is the center and ro is the radius of the inclusion.
References
1. Osher, S., Sethian, J.A. (1988) "Fronts propagating with curvature dependent speed: Algorithms based on Hamilton-Jacobi formulations," Journal of Computational Physics, 79, 12-49.
2. Adalsteinsson, D., Sethian, J.A. (1995) "A fast level set method for propagating interfaces," Journal of Computational Physics, 118, 269-277.
3. Stolarska, M., Chopp, D.L., Moes, N., Belytschko, T. (2001) "Modelling crack growth by level sets in the extended finite element method," International Journal for Numerical Methods in Engineering, 51, 943-960.
4. Sukumar, N., Chopp, D.L., Moes, N., Belytschko, T. (2001) "Modeling holes and inclusions by level sets in the extended finite-element method," Computer Methods in Applied Mechanics and Engineering, 190, 6183-6200. | 646 | 2,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-38 | latest | en | 0.929245 |
https://derekwadsworth.com/experimental-probability-worksheet/ | 1,656,980,678,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104506762.79/warc/CC-MAIN-20220704232527-20220705022527-00112.warc.gz | 248,146,584 | 6,480 | This 1-page, 11-question worksheet is a an excellent way because that students to practice finding the theoretical and also experimental probabilities of straightforward events! Please view the preview document to identify if this product is appropriate for her students!An answer an essential is included.This activity is aligned to CCSS 7.
You are watching: Experimental probability worksheet
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Teachers Pay teachers is an virtual marketplace whereby teachers buy and sell initial educational materials. | 1,416 | 7,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-27 | latest | en | 0.928019 |
https://teacheradvisor.org/2/activities/chunk_2-1-6-3 | 1,618,406,928,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077818.23/warc/CC-MAIN-20210414125133-20210414155133-00057.warc.gz | 632,185,092 | 7,440 | Fluency Activity
Take Out Ten
Lesson 6. Unit 1. Grade 2 EngageNY
EngageNY3 min(s)
This Fluency Activity is a part of the Lesson 6, Unit 1, Grade 2. Taking out 10 prepares students for subtracting a single-digit from a two-digit number where there are not enough ones. Let's take out 10 from each number. I say 30. You draw a number bond for 30 with parts 20 and 10. Show the ten on the right. | 115 | 396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-17 | latest | en | 0.839031 |
https://www.mrexcel.com/board/threads/reverse-a-string.222979/ | 1,716,800,213,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059039.52/warc/CC-MAIN-20240527083011-20240527113011-00452.warc.gz | 782,605,883 | 21,396 | # Reverse a string
#### BrianM
##### Well-known Member
Is there a function that can reverse the contents of a cell?
i.e. A1=Cat ...result = tac
Thanks,
Brian
### Excel Facts
Pivot Table Drill Down
Double-click any number in a pivot table to create a new report showing all detail rows that make up that number
Is there a function that can reverse the contents of a cell?
i.e. A1=Cat ...result = tac
Thanks,
Brian
ASAP Utilities has a function that does just that,
www.asap-utilities.com
Or a User Defined function
Code:
``````Function reverse(S As String)
Dim Temp As String, I As Integer, n As Integer
n = Len(S)
For I = n To 1 Step -1
Temp = Temp & Mid(S, I, 1)
Next I
reverse = Temp & Right(S, Len(S) - n)
End Function``````
ASAP-Utilities is better though unless you are going to be doing this a lot.
Thanks to you both.
Hi,
there is an inbuiltfunction
might be a more recent one
Code:
``MsgBox StrReverse("cat")``
kind regards,
Erik
Good evening BrianM
This would do the trick :
Code:
``````Function Reverse(Backwards As String)
Dim Length, Pos As Integer
Reverse = ""
Length = Len(Backwards)
For Pos = Length To 1 Step -1
Reverse = Reverse & Mid(Backwards, Pos, 1)
Next Pos
End Function``````
My add-in, available from the link below has this as a function, so once installed =REVERSE(A1) would show tac (if A1 contained "cat")
HTH
DominicB
Code:
``````Function reverse(S As String)
reverse = StrReverse(S)
End Function``````
so no need to loop
as mentioned, check if it works for your version
Using formulas, (not too pretty though):-
Book1
ABCDEFGHIJKLMNOPQRSTUVWXYZAAABAC
1ToddBardoniinodraB ddoTTTTTTTTTTTTTTTinodraBddoT
Sheet1
In C1:AB1
Code:
``=LEFT(RIGHT(Sheet1!\$A\$1,COLUMN(1:\$256)),1)``
Confirmed w/ Ctrl+Shift+Enter
In AC1
Code:
``=LEFT(C1&D1&E1&F1&G1&H1&I1&J1&K1&L1&M1&N1&O1&P1&Q1&R1&S1&T1&U1&V1&W1&X1&Y1&Z1&AA1&AB1,LEN(\$A\$1))``
Hi Erik
That "StrReverse" commands a handy thing to know. It would be interesting to know when it was introduced. As far back as XL2000...?
DominicB
Domonic,
I'll check it out on an elder PC with XL2000.
Almost sure it wasn't available in XL97
best regards,
Erik
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Lapis Lazuli | Level 10
Use Multiple Colors in a Graph
I am attempting to make a "waterfall" plot with the NEEDLE interpolation in PROC GPLOT. My plot is divided into two different groups. I wanted the positive group to be blue and the negative group to be red. Here is what I have attempted:
axis1 label=(angle=90 'Best Percent Change in Lesion Size') order=(-100 to 150 by 50) minor=(n=4);
axis2 label=(' ') value=none major=none minor=none;
proc gplot data=waterfall;
plot ploty1*n ploty2*n / overlay vaxis=axis1 haxis=axis2;
symbol1 interpol=Needle value=none width=4 color=blue;
symbol2 interpol=Needle value=none width=4 color=red;
run;
I am not at all familiar with annotations, but I am sure I could do something with that. I just have no idea where to start.
This is not my full data, but here is an idea of how it looks:
brpercentposnegnploty1ploty2
2.564Positive12.564
2.083Positive22.083
1.333Positive31.333
0Positive40
-5.263Positive5-.263
-30Positive6-30
-31.429Positive7-31.429
-36.842Positive8-36.842
-40Positive9-40
-50Positive10-50
150Negative11150
142.857Negative12142.857
78.152Negative1378.152
58.065Negative1458.065
52.174Negative1552.174
-3.704Negative16-3.704
-4Negative17-4
-4.762Negative18-4.762
-7.143Negative19-7.143
-18.75Negative20-18.75
1 ACCEPTED SOLUTION
Accepted Solutions
Meteorite | Level 14
Use Multiple Colors in a Graph
Here's an example where I've used Proc Gchart and Annotate to create a waterfall chart - I think you can re-use this technique to get the graph you're wanting...
http://robslink.com/SAS/democd41/waterfall_anno.htm
http://robslink.com/SAS/democd41/waterfall_anno_info.htm
1 REPLY 1
Meteorite | Level 14
Use Multiple Colors in a Graph
Here's an example where I've used Proc Gchart and Annotate to create a waterfall chart - I think you can re-use this technique to get the graph you're wanting...
http://robslink.com/SAS/democd41/waterfall_anno.htm
http://robslink.com/SAS/democd41/waterfall_anno_info.htm
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http://mathhelpforum.com/calculus/53718-understanding-range-functions-2-variables.html | 1,481,341,224,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542938.92/warc/CC-MAIN-20161202170902-00389-ip-10-31-129-80.ec2.internal.warc.gz | 169,574,031 | 9,960 | # Thread: Understanding range in functions of 2 variables
1. ## Understanding range in functions of 2 variables
I am having trouble understanding the range in f(x,y)=4x^2+9y^2. The answer is z>=0 but my book is not helping me understand how to find this. Can someone please help explain this to me? Thank you.
2. Originally Posted by cowboys111
I am having trouble understanding the range in f(x,y)=4x^2+9y^2. The answer is z>=0 but my book is not helping me understand how to find this. Can someone please help explain this to me? Thank you.
we have $z = 4x^2 + 9y^2$
now, $4x^2 \ge 0$ for all $x$ and $9y^2 \ge 0$ for all $y$. 0 is the minimum for both of these
thus, $z \ge 0 + 0 = 0$ | 215 | 692 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2016-50 | longest | en | 0.933106 |
https://www.sciensation.org/hands-on_experiments/e5066p_balancingStick.html | 1,519,049,697,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812665.41/warc/CC-MAIN-20180219131951-20180219151951-00263.warc.gz | 922,888,473 | 6,682 | • What is easier?
• That depends on where the center of gravity lies.
• Because this determines the rotational inertia of the system.
www.sciensation.org | Ciênsação hands-on experiments are published as Open Educational resources under a Creative Commons Attribution-ShareAlike 4.0 International License.
Physics Game
age: 14 – 17
# Balancing stick
Newton't first law of inertia also applies to rotations – and comes in handy when balancing a stick on a fingertip.
Learning objective
Experiencing the effect of rotational inertia.
Sticks (e.g. meat skewers or rulers)
Weights that can be attached to the stick (e.g. a fruit or modeling clay)
Preparation
Prepare 3 or 4 identical sticks (e.g. meat skewers or rulers, the longer the better) by attaching weights near one of the ends.
Student Task
1. Try to balance the stick on your fingertip. Who can balance it the longest?
2. Why is it easier to balance the stick when the weight is near the top?
Guiding Questions (if needed)
Where is the center of gravity?
› You can easily find it by balancing the object horizontally.
Around what point does the center of gravity rotate?
› Around the fingertip.
Why does the radius of the rotation matter?
› The larger the radius, the larger the angular momentum and thus the slower the stick tilts.
Findings
The further away the center of gravity is located from the pivot point (here the fingertip), the greater the rotational inertia, and the slower the stick tilts – allowing more time to adjust and thus making it easier to maintain balance. | 341 | 1,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2018-09 | longest | en | 0.876038 |
https://learn.careers360.com/engineering/question-how-to-solve-this-problem-which-of-the-following-is-true/ | 1,723,161,903,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640741453.47/warc/CC-MAIN-20240808222119-20240809012119-00673.warc.gz | 274,087,296 | 35,358 | #### Which of the following is true ? Option 1) is continous at x= 0 Option 2) is continous at x= 0 Option 3) is continous at x= 1([.]=G.I.F) Option 4) is continous at x= 1.5([.]= G.I.F)
As we have learned
Continuity at a point -
A function f(x) is said to be continuous at x = a in its domain if
1. f(a) is defined : at x = a.
of f(x) at x = a exists from left and right.
-
In(A) , f(x) is not defined at x=0
In (B) , f(x) is not defined at x=0 and also LHLRHL
In (C) , f(x) is defined at x= 1, f(1) =1
But f(1) LHL, RHL all are not same
In (D) f(1.5)=
Option 1)
is continous at x= 0
Option 2)
is continous at x= 0
Option 3)
is continous at x= 1([.]=G.I.F)
Option 4)
is continous at x= 1.5([.]= G.I.F) | 299 | 741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-33 | latest | en | 0.776284 |
https://www.quantumstudy.com/unpolarized-light-falls-on-two-sheets-placed-one-on-top-of-the-other-what-must-be-the-angle-between-the-characteristic-direction-of-the-sheets-if-the-intensity-of-the-transmitted-light-is-one-third-o/ | 1,696,051,453,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510603.89/warc/CC-MAIN-20230930050118-20230930080118-00651.warc.gz | 1,051,172,805 | 49,255 | # Unpolarized light falls on two sheets placed one on top of the other. What must be the angle between the characteristic direction of the sheets if the intensity of the transmitted light is one third of intensity of the incident beam ?
Q: Unpolarized light falls on two sheets placed one on top of the other. What must be the angle between the characteristic direction of the sheets if the intensity of the transmitted light is one third of intensity of the incident beam ?
Sol: Intensity of the light transmitted through the first polarized I1 = I0/2 , where I0 is the intensity of the incident unpolarized light. Intensity of the light transmitted through the second polarize is I2 = I1 cos2θ
θ is the angle between the characteristic directions of the polarizer sheets.
But I2 = I0/ 3 (given)
∴ I2 = I1 cos2θ = (I0/2) cos2 θ
I0/ 3 = (I0/2) cos2 θ
∴ cos2 θ = 2/3
⇒ $\large \theta = cos^{-1}\sqrt{\frac{2}{3}}$ | 246 | 920 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2023-40 | latest | en | 0.8877 |
http://mathhelpforum.com/calculus/111275-double-integrals-print.html | 1,529,562,635,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864039.24/warc/CC-MAIN-20180621055646-20180621075646-00176.warc.gz | 219,964,792 | 2,810 | # double integrals!
• Oct 29th 2009, 06:08 PM
collegestudent321
double integrals!
hi guys, i have a question that im pretty sure im doing right but i keep gettin the wrong answer. it says:
Find mass and Center of Mass for the following:
Dis bounded by the parabola x = y^2 and the line y = x - 2 with density(x,y) = 3. for the mass, i did:
m = (integral from x = 0 to x = 4)(integral from y = x-2 to y = x^(1/2) of 3 dydx
I am getting m = 16, but my book says m = 27/2 = 13.5
Could you help me find out what my mistake is??? Thank you in advance!
• Nov 1st 2009, 06:46 AM
Media_Man
Bounds of integration
First, you have the wrong bounds of integration. Second, you are integrating with respect to x instead of y. Try drawing the two graphs on a piece of paper and look carefully at the situation.
$\displaystyle x_1=y^2, x_2=y+2$ has intersection points $\displaystyle (1,-1),(2,4)$. So the integral is $\displaystyle A=\int_{-1}^2 y+2-y^2 dy$. Multiplying this result by the constant density $\displaystyle \rho=3$ does indeed get you a correct answer. | 324 | 1,059 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-26 | latest | en | 0.905332 |
http://www.jiskha.com/display.cgi?id=1208552908 | 1,495,717,015,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608067.23/warc/CC-MAIN-20170525121448-20170525141448-00570.warc.gz | 555,577,651 | 4,256 | chemistry
posted by on .
Question:A solution of AgNO3 (45 mL/0.45 M) was mixed with solution of NaCl (85 mL/1.35 x 10-2 M)
a) Calculate the ion product of the potential precipitate.
mol AgNO3 = M x L
= 0.45 M x 0.045 M
= 0.02025
mol NaCl = M x L
= 0.0135 M x 0.085 L
= 0.0011475
(Ag^+) = mols / total volume
= 0.02025 / 0.130 L
= 0.1558
total volume = 85 mL + 45 mL
= 130 mL
= 0.130 L
(Cl^-) = mols / total volume
= 0.0011475 / 0.130 L
= 0.00883
Ion product = [Ag^+][Cl^-]
= [0.1558][0.00883]
= 0.001375
• chemistry - ,
It looks ok to me. One point I want to make here. Technically, the Ksp of a salt, in this case, AgCl, cannot be exceeded (unless a supersaturated solution is formed). That is, (Ag^+)(Cl^-) can not be larger than Ksp which is about 1.8 x 10^-10. That's why what you have done is called the ion product. Since the ion product exceeds Ksp, you know a ppt will occur. How much AgCl will ppt. Enough AgCl will come out of solution so that the final (Ag^+) x the final (Cl^-) will be equal to 1.8 x 10^-10. In a saturated solution, (Ag^+) = (Cl^-) = 1.34 x 10^-5 molar so most will ppt. The point I want to make is that the ion product thing is an artificial one in that Ksp can't be exceeded. But the ion product is a way of comparing Ksp and knowing if a ppt will occur.
• chemistry - ,
Thanks for the tip | 463 | 1,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2017-22 | latest | en | 0.881267 |
http://www.atharvnadkarni.com/blog/author/lthmath/ | 1,726,480,652,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651682.69/warc/CC-MAIN-20240916080220-20240916110220-00408.warc.gz | 32,201,524 | 9,026 | # Proof Writing (Part 1): An introduction to a couple types of proof.
Hello, and welcome back, and today we’re going to be pausing our discussion of Pascal’s triangle temporarily, and discussing some proof writing! (Don’t worry, this…
# Pascal’s Triangle (Part 3):
Hello, and welcome to the 3rd episode in this 5 part series! In this blog we will be formally writing the Hockey-stick Principle showcased in…
# Pascal’s Triangle (Part 2)
Hello everyone! Welcome back to my blog series on Pascal’s Triangle. Today we will be looking at the first pattern covered in the last blog–triangular…
# Pascal’s Triangle (Part 1)
In this blog, we discus some of the more basic properties in Pascal’s triangle, and lay a foundation for the future blogs in this series!
# Modular Arithmetic Part 5 (Finale)
The pinnacle of this series of modular arithmetic, we prove a theorem that ties everything together!
# Modular Arithmetic Part 4 (Multiplication tables and inverses)
In the penultimate episode of modular arithmetic, we discuss the relation between inverses, and multiplication tables, and set the scene for part 5!
# Modular Arithmetic: Part 3 (Proving stuff)
Explore the world of proofs while learning about the inner workings of modular arithmetic!
# Modular Arithmetic Part 2 (Properties and equations)
Delve more into the realm of modular arithmetic as we learn about doing operations in it! | 304 | 1,403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-38 | latest | en | 0.851878 |
https://nursingcoursework.org/1when-a-balloon-rises-in-the-atmosphere-it-expands-as-it-goes-up-what-factors-determine-by-what-factor-it-expands-and-when-it-will-break-apart/ | 1,603,641,050,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889574.66/warc/CC-MAIN-20201025154704-20201025184704-00135.warc.gz | 455,147,885 | 20,534 | # 1)When a balloon rises in the atmosphere it expands as it goes up. What factors determine by what factor it expands and when it will break apart?
October 18, 2020
###### create a short research proposal for an ethnographic study that answers questions only answerable through qualitative research.
October 18, 2020
1)When a balloon rises in the atmosphere it expands as it goes up. What factors determine by what factor it expands and when it will break apart? Assume the balloon is sealed and does not lose gas during its flight. Select all that apply.
A-The choice of gas (say hydrogen or helium) for filling the balloon.
B-The temperature profile of Earth’s atmosphere.
C-Decrease of Earth’s atmospheric pressure with altitude
D-Hooke’s Law elasticity of the material of the balloon
2)The average energy of an atom or molecule in a gas is 3/2 kT where k is the Boltzmann Constant, and T is the temperature in kelvins. If you were to look at all the atoms in a snapshot of the gas and sort them out by energy, what is the most probable energy you would find for an atom?
A-1/2 kT
B-3/2 kT
C-2 kT
D-8 k T / π
3)Carbon, nitrogen, and oxygen have 6, 7 and 8 protons in their nucleus. How many neutrons are in their nucleus as well, for the normal stable isotopes of these elements.
A- 6, 7, and 8
B- 7, 8, and 9
C- 12, 14, and 16
D- none
4)In a gas that is a mixture of water (two H and one O) and carbon dioxide (one C and two O), if the water has average velocity of 559 m/s, what is the average velocity of the carbon dioxide molecule?x
##### . WITH nursingcoursework.org AND GET AN AMAZING DISCOUNT!
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1)When a balloon rises in the atmosphere it expands as it goes up. What factors determine by what factor it expands and when it will break apart? was first posted on October 18, 2020 at 10:58 am. | 532 | 2,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-45 | latest | en | 0.925071 |
https://number.academy/1005613 | 1,708,627,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.13/warc/CC-MAIN-20240222161802-20240222191802-00532.warc.gz | 446,571,738 | 12,257 | # Number 1005613 facts
The odd number 1,005,613 is spelled 🔊, and written in words: one million, five thousand, six hundred and thirteen, approximately 1.0 million. The ordinal number 1005613rd is said 🔊 and written as: one million, five thousand, six hundred and thirteenth. The meaning of the number 1005613 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1005613. What is 1005613 in computer science, numerology, codes and images, writing and naming in other languages
## What is 1,005,613 in other units
The decimal (Arabic) number 1005613 converted to a Roman number is (M)(V)DCXIII. Roman and decimal number conversions.
#### Weight conversion
1005613 kilograms (kg) = 2216974.4 pounds (lbs)
1005613 pounds (lbs) = 456143.1 kilograms (kg)
#### Length conversion
1005613 kilometers (km) equals to 624859 miles (mi).
1005613 miles (mi) equals to 1618378 kilometers (km).
1005613 meters (m) equals to 3299216 feet (ft).
1005613 feet (ft) equals 306515 meters (m).
1005613 centimeters (cm) equals to 395910.6 inches (in).
1005613 inches (in) equals to 2554257.0 centimeters (cm).
#### Temperature conversion
1005613° Fahrenheit (°F) equals to 558656.1° Celsius (°C)
1005613° Celsius (°C) equals to 1810135.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
1005613 seconds equals to 1 week, 4 days, 15 hours, 20 minutes, 13 seconds
1005613 minutes equals to 2 years, 3 weeks, 5 days, 8 hours, 13 minutes
### Codes and images of the number 1005613
Number 1005613 morse code: .---- ----- ----- ..... -.... .---- ...--
Sign language for number 1005613:
Number 1005613 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 1005613
### Multiplications
#### Multiplication table of 1005613
1005613 multiplied by two equals 2011226 (1005613 x 2 = 2011226).
1005613 multiplied by three equals 3016839 (1005613 x 3 = 3016839).
1005613 multiplied by four equals 4022452 (1005613 x 4 = 4022452).
1005613 multiplied by five equals 5028065 (1005613 x 5 = 5028065).
1005613 multiplied by six equals 6033678 (1005613 x 6 = 6033678).
1005613 multiplied by seven equals 7039291 (1005613 x 7 = 7039291).
1005613 multiplied by eight equals 8044904 (1005613 x 8 = 8044904).
1005613 multiplied by nine equals 9050517 (1005613 x 9 = 9050517).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 1005613
Half of 1005613 is 502806,5 (1005613 / 2 = 502806,5 = 502806 1/2).
One third of 1005613 is 335204,3333 (1005613 / 3 = 335204,3333 = 335204 1/3).
One quarter of 1005613 is 251403,25 (1005613 / 4 = 251403,25 = 251403 1/4).
One fifth of 1005613 is 201122,6 (1005613 / 5 = 201122,6 = 201122 3/5).
One sixth of 1005613 is 167602,1667 (1005613 / 6 = 167602,1667 = 167602 1/6).
One seventh of 1005613 is 143659 (1005613 / 7 = 143659).
One eighth of 1005613 is 125701,625 (1005613 / 8 = 125701,625 = 125701 5/8).
One ninth of 1005613 is 111734,7778 (1005613 / 9 = 111734,7778 = 111734 7/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
1005613
#### Is Prime?
The number 1005613 is not a prime number. The closest prime numbers are 1005593, 1005617.
#### Factorization and factors (dividers)
The prime factors of 1005613 are 7 * 19 * 7561
The factors of 1005613 are 1, 7, 19, 133, 7561, 52927, 143659, 1005613.
Total factors 8.
Sum of factors 1209920 (204307).
#### Powers
The second power of 10056132 is 1.011.257.505.769.
The third power of 10056133 is 1.016.933.694.148.881.408.
#### Roots
The square root √1005613 is 1002,802573.
The cube root of 31005613 is 100,186751.
#### Logarithms
The natural logarithm of No. ln 1005613 = loge 1005613 = 13,821108.
The logarithm to base 10 of No. log10 1005613 = 6,002431.
The Napierian logarithm of No. log1/e 1005613 = -13,821108.
### Trigonometric functions
The cosine of 1005613 is -0,186069.
The sine of 1005613 is 0,982537.
The tangent of 1005613 is -5,280484.
### Properties of the number 1005613
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 1005613 in Computer Science
Code typeCode value
1005613 Number of bytes982.0KB
Unix timeUnix time 1005613 is equal to Monday Jan. 12, 1970, 3:20:13 p.m. GMT
IPv4, IPv6Number 1005613 internet address in dotted format v4 0.15.88.45, v6 ::f:582d
1005613 Decimal = 11110101100000101101 Binary
1005613 Decimal = 1220002102221 Ternary
1005613 Decimal = 3654055 Octal
1005613 Decimal = F582D Hexadecimal (0xf582d hex)
1005613 BASE64MTAwNTYxMw==
1005613 MD5261035c29e2684479623dd32b5b44f6d
1005613 SHA256ca0130551da3169938b89220efef61f44f3e6dd153ee9a9ddb29ae418b626b7d
1005613 SHA3846de090e38dccfa9f472a0a059d2edce243885a3053b111600fdd903fc9170f7fd6890151fd32442bb9829c61bb0065c8
More SHA codes related to the number 1005613 ...
If you know something interesting about the 1005613 number that you did not find on this page, do not hesitate to write us here.
## Numerology 1005613
### Character frequency in the number 1005613
Character (importance) frequency for numerology.
Character: Frequency: 1 2 0 2 5 1 6 1 3 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 1005613, the numbers 1+0+0+5+6+1+3 = 1+6 = 7 are added and the meaning of the number 7 is sought.
## № 1,005,613 in other languages
How to say or write the number one million, five thousand, six hundred and thirteen in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.005.613) un millón cinco mil seiscientos trece German: 🔊 (Nummer 1.005.613) eine Million fünftausendsechshundertdreizehn French: 🔊 (nombre 1 005 613) un million cinq mille six cent treize Portuguese: 🔊 (número 1 005 613) um milhão e cinco mil, seiscentos e treze Hindi: 🔊 (संख्या 1 005 613) दस लाख, पाँच हज़ार, छः सौ, तेरह Chinese: 🔊 (数 1 005 613) 一百万五千六百一十三 Arabian: 🔊 (عدد 1,005,613) واحد مليون و خمسة آلاف و ستمائة و ثلاثة عشر Czech: 🔊 (číslo 1 005 613) milion pět tisíc šestset třináct Korean: 🔊 (번호 1,005,613) 백만 오천육백십삼 Dutch: 🔊 (nummer 1 005 613) een miljoen vijfduizendzeshonderddertien Japanese: 🔊 (数 1,005,613) 百万五千六百十三 Indonesian: 🔊 (jumlah 1.005.613) satu juta lima ribu enam ratus tiga belas Italian: 🔊 (numero 1 005 613) un milione e cinquemilaseicentotredici Norwegian: 🔊 (nummer 1 005 613) en million, fem tusen, seks hundre og tretten Polish: 🔊 (liczba 1 005 613) milion pięć tysięcy sześćset trzynaście Russian: 🔊 (номер 1 005 613) один миллион пять тысяч шестьсот тринадцать Turkish: 🔊 (numara 1,005,613) birmilyonbeşbinaltıyüzonüç Thai: 🔊 (จำนวน 1 005 613) หนึ่งล้านห้าพันหกร้อยสิบสาม Ukrainian: 🔊 (номер 1 005 613) один мільйон п'ять тисяч шістсот тринадцять Vietnamese: 🔊 (con số 1.005.613) một triệu năm nghìn sáu trăm mười ba Other languages ...
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https://www.coursehero.com/file/6164448/MP-and-MT/ | 1,493,224,065,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121453.27/warc/CC-MAIN-20170423031201-00358-ip-10-145-167-34.ec2.internal.warc.gz | 888,209,994 | 71,860 | # MP and MT - Todays Topics (11/9) I. a. Two Common Valid...
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3/4/11 Today’s Topics (11/9) I. Two Common Valid Argument Forms a. Modus Ponens b. Modus Tollens II. A Point about Validity III. Some Other Rules of Inference
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3/4/11 II. Two Common Valid Argument Forms “Modus Ponens” names a particular argument form: p ⊃ q p In arguments of this form, one of the premises is a conditional. The other premise affirms the antecedent. The conclusion affirms the consequent. “Modus Ponens” is sometimes called “ Affirming the Antecedent .”
3/4/11 Proof that Modus Ponens is a Valid Argument Form p q p ⊃ q p q T T T T T T F F T F F T T F T F F T F F OK
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3/4/11 Implication Every argument that is an instance of a valid argument form Since the following arguments are all instances of Modus Ponens, they are all valid. 1. If James drove drunk, he broke the law. 2. James drove drunk. ∴ 3. James broke the law. 1. If James did not break the law, he did not drive drunk. 2. James did not break the law. ∴ 3. James did not drive drunk. 1. If the president is a democrat and most of the senators are democrats, then the Republicans are not happy. 2. The president is a democrat and most of the senators are democrats. ∴ 3. The Republicans are not happy.
Modus Ponens as a Rule of Inference From premises of the form “If p then q” and “p” you can validly infer “q.” Suppose you believe that a. If your wife is not home, she is having an affair; and b. Your wife is not home. You can validly infer that your wife is having an affair.
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## This note was uploaded on 03/03/2011 for the course PHIL 100 taught by Professor Forte during the Spring '08 term at Bridgewater State University.
### Page1 / 17
MP and MT - Todays Topics (11/9) I. a. Two Common Valid...
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Ask a homework question - tutors are online | 614 | 2,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-17 | longest | en | 0.892351 |
http://www.netlib.org/lapack/explore-html/d4/d19/zgeqr_8f_a8ae88d8fdedeebfac10501b91406241b.html | 1,556,002,416,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578593360.66/warc/CC-MAIN-20190423054942-20190423080942-00300.warc.gz | 275,971,927 | 7,123 | LAPACK 3.8.0 LAPACK: Linear Algebra PACKage
## ◆ zgeqr()
subroutine zgeqr ( integer M, integer N, complex*16, dimension( lda, * ) A, integer LDA, complex*16, dimension( * ) T, integer TSIZE, complex*16, dimension( * ) WORK, integer LWORK, integer INFO )
Purpose:
ZGEQR computes a QR factorization of an M-by-N matrix A.
Parameters
[in] M ``` M is INTEGER The number of rows of the matrix A. M >= 0.``` [in] N ``` N is INTEGER The number of columns of the matrix A. N >= 0.``` [in,out] A ``` A is COMPLEX*16 array, dimension (LDA,N) On entry, the M-by-N matrix A. On exit, the elements on and above the diagonal of the array contain the min(M,N)-by-N upper trapezoidal matrix R (R is upper triangular if M >= N); the elements below the diagonal are used to store part of the data structure to represent Q.``` [in] LDA ``` LDA is INTEGER The leading dimension of the array A. LDA >= max(1,M).``` [out] T ``` T is COMPLEX*16 array, dimension (MAX(5,TSIZE)) On exit, if INFO = 0, T(1) returns optimal (or either minimal or optimal, if query is assumed) TSIZE. See TSIZE for details. Remaining T contains part of the data structure used to represent Q. If one wants to apply or construct Q, then one needs to keep T (in addition to A) and pass it to further subroutines.``` [in] TSIZE ``` TSIZE is INTEGER If TSIZE >= 5, the dimension of the array T. If TSIZE = -1 or -2, then a workspace query is assumed. The routine only calculates the sizes of the T and WORK arrays, returns these values as the first entries of the T and WORK arrays, and no error message related to T or WORK is issued by XERBLA. If TSIZE = -1, the routine calculates optimal size of T for the optimum performance and returns this value in T(1). If TSIZE = -2, the routine calculates minimal size of T and returns this value in T(1).``` [out] WORK ``` (workspace) COMPLEX*16 array, dimension (MAX(1,LWORK)) On exit, if INFO = 0, WORK(1) contains optimal (or either minimal or optimal, if query was assumed) LWORK. See LWORK for details.``` [in] LWORK ``` LWORK is INTEGER The dimension of the array WORK. If LWORK = -1 or -2, then a workspace query is assumed. The routine only calculates the sizes of the T and WORK arrays, returns these values as the first entries of the T and WORK arrays, and no error message related to T or WORK is issued by XERBLA. If LWORK = -1, the routine calculates optimal size of WORK for the optimal performance and returns this value in WORK(1). If LWORK = -2, the routine calculates minimal size of WORK and returns this value in WORK(1).``` [out] INFO ``` INFO is INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value```
Further Details
The goal of the interface is to give maximum freedom to the developers for creating any QR factorization algorithm they wish. The triangular (trapezoidal) R has to be stored in the upper part of A. The lower part of A and the array T can be used to store any relevant information for applying or constructing the Q factor. The WORK array can safely be discarded after exit.
Caution: One should not expect the sizes of T and WORK to be the same from one LAPACK implementation to the other, or even from one execution to the other. A workspace query (for T and WORK) is needed at each execution. However, for a given execution, the size of T and WORK are fixed and will not change from one query to the next.
Further Details particular to this LAPACK implementation:
These details are particular for this LAPACK implementation. Users should not take them for granted. These details may change in the future, and are unlikely not true for another LAPACK implementation. These details are relevant if one wants to try to understand the code. They are not part of the interface.
In this version,
T(2): row block size (MB) T(3): column block size (NB) T(6:TSIZE): data structure needed for Q, computed by ZLATSQR or ZGEQRT
Depending on the matrix dimensions M and N, and row and column block sizes MB and NB returned by ILAENV, ZGEQR will use either ZLATSQR (if the matrix is tall-and-skinny) or ZGEQRT to compute the QR factorization.
Definition at line 162 of file zgeqr.f.
162 *
163 * -- LAPACK computational routine (version 3.7.0) --
164 * -- LAPACK is a software package provided by Univ. of Tennessee, --
165 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd. --
166 * December 2016
167 *
168 * .. Scalar Arguments ..
169 INTEGER info, lda, m, n, tsize, lwork
170 * ..
171 * .. Array Arguments ..
172 COMPLEX*16 a( lda, * ), t( * ), work( * )
173 * ..
174 *
175 * =====================================================================
176 *
177 * ..
178 * .. Local Scalars ..
179 LOGICAL lquery, lminws, mint, minw
180 INTEGER mb, nb, mintsz, nblcks
181 * ..
182 * .. External Functions ..
183 LOGICAL lsame
184 EXTERNAL lsame
185 * ..
186 * .. External Subroutines ..
187 EXTERNAL zlatsqr, zgeqrt, xerbla
188 * ..
189 * .. Intrinsic Functions ..
190 INTRINSIC max, min, mod
191 * ..
192 * .. External Functions ..
193 INTEGER ilaenv
194 EXTERNAL ilaenv
195 * ..
196 * .. Executable Statements ..
197 *
198 * Test the input arguments
199 *
200 info = 0
201 *
202 lquery = ( tsize.EQ.-1 .OR. tsize.EQ.-2 .OR.
203 \$ lwork.EQ.-1 .OR. lwork.EQ.-2 )
204 *
205 mint = .false.
206 minw = .false.
207 IF( tsize.EQ.-2 .OR. lwork.EQ.-2 ) THEN
208 IF( tsize.NE.-1 ) mint = .true.
209 IF( lwork.NE.-1 ) minw = .true.
210 END IF
211 *
212 * Determine the block size
213 *
214 IF( min( m, n ).GT.0 ) THEN
215 mb = ilaenv( 1, 'ZGEQR ', ' ', m, n, 1, -1 )
216 nb = ilaenv( 1, 'ZGEQR ', ' ', m, n, 2, -1 )
217 ELSE
218 mb = m
219 nb = 1
220 END IF
221 IF( mb.GT.m .OR. mb.LE.n ) mb = m
222 IF( nb.GT.min( m, n ) .OR. nb.LT.1 ) nb = 1
223 mintsz = n + 5
224 IF( mb.GT.n .AND. m.GT.n ) THEN
225 IF( mod( m - n, mb - n ).EQ.0 ) THEN
226 nblcks = ( m - n ) / ( mb - n )
227 ELSE
228 nblcks = ( m - n ) / ( mb - n ) + 1
229 END IF
230 ELSE
231 nblcks = 1
232 END IF
233 *
234 * Determine if the workspace size satisfies minimal size
235 *
236 lminws = .false.
237 IF( ( tsize.LT.max( 1, nb*n*nblcks + 5 ) .OR. lwork.LT.nb*n )
238 \$ .AND. ( lwork.GE.n ) .AND. ( tsize.GE.mintsz )
239 \$ .AND. ( .NOT.lquery ) ) THEN
240 IF( tsize.LT.max( 1, nb*n*nblcks + 5 ) ) THEN
241 lminws = .true.
242 nb = 1
243 mb = m
244 END IF
245 IF( lwork.LT.nb*n ) THEN
246 lminws = .true.
247 nb = 1
248 END IF
249 END IF
250 *
251 IF( m.LT.0 ) THEN
252 info = -1
253 ELSE IF( n.LT.0 ) THEN
254 info = -2
255 ELSE IF( lda.LT.max( 1, m ) ) THEN
256 info = -4
257 ELSE IF( tsize.LT.max( 1, nb*n*nblcks + 5 )
258 \$ .AND. ( .NOT.lquery ) .AND. ( .NOT.lminws ) ) THEN
259 info = -6
260 ELSE IF( ( lwork.LT.max( 1, n*nb ) ) .AND. ( .NOT.lquery )
261 \$ .AND. ( .NOT.lminws ) ) THEN
262 info = -8
263 END IF
264 *
265 IF( info.EQ.0 ) THEN
266 IF( mint ) THEN
267 t( 1 ) = mintsz
268 ELSE
269 t( 1 ) = nb*n*nblcks + 5
270 END IF
271 t( 2 ) = mb
272 t( 3 ) = nb
273 IF( minw ) THEN
274 work( 1 ) = max( 1, n )
275 ELSE
276 work( 1 ) = max( 1, nb*n )
277 END IF
278 END IF
279 IF( info.NE.0 ) THEN
280 CALL xerbla( 'ZGEQR', -info )
281 RETURN
282 ELSE IF( lquery ) THEN
283 RETURN
284 END IF
285 *
286 * Quick return if possible
287 *
288 IF( min( m, n ).EQ.0 ) THEN
289 RETURN
290 END IF
291 *
292 * The QR Decomposition
293 *
294 IF( ( m.LE.n ) .OR. ( mb.LE.n ) .OR. ( mb.GE.m ) ) THEN
295 CALL zgeqrt( m, n, nb, a, lda, t( 6 ), nb, work, info )
296 ELSE
297 CALL zlatsqr( m, n, mb, nb, a, lda, t( 6 ), nb, work,
298 \$ lwork, info )
299 END IF
300 *
301 work( 1 ) = max( 1, nb*n )
302 *
303 RETURN
304 *
305 * End of ZGEQR
306 *
integer function ilaenv(ISPEC, NAME, OPTS, N1, N2, N3, N4)
ILAENV
Definition: tstiee.f:83
subroutine zlatsqr(M, N, MB, NB, A, LDA, T, LDT, WORK, LWORK, INFO)
Definition: zlatsqr.f:151
subroutine xerbla(SRNAME, INFO)
XERBLA
Definition: xerbla.f:62
logical function lsame(CA, CB)
LSAME
Definition: lsame.f:55
subroutine zgeqrt(M, N, NB, A, LDA, T, LDT, WORK, INFO)
ZGEQRT
Definition: zgeqrt.f:143
Here is the call graph for this function:
Here is the caller graph for this function: | 2,708 | 8,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-18 | latest | en | 0.733764 |
http://www.onlinemathlearning.com/division-area-model-remainders.html | 1,493,605,700,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917126538.54/warc/CC-MAIN-20170423031206-00205-ip-10-145-167-34.ec2.internal.warc.gz | 643,956,166 | 9,396 | # Division with Remainders using the Area Model
Videos and solutions to help Grade 4 students learn how to solve division problems with remainders using the area model.
Common Core Standards: 4.NBT.6, 4.OA.3
New York State Common Core Math Grade 4, Module 3, Lesson 21
NYS Math Module 3, Grade 4, Lesson 21 Problem Set
Problem 1: Solve 37 ÷ 2 using an area model. Use long division and distributive property to record your work.
Problem 2. Solve 76 ÷ 3 using an area model. Use long division and the distributive property to record your work.
3. Carolina solved the following division problem by drawing an area model.
a. What division problem did she solve?
b. Show how Carolina’s model can be represented using the distributive property. NYS Math Module 3 Grade 4 Lesson 21 Problem Set
Problem 1 - Problem 3
Solve the following problems using the area model. Support the area model with long division or the distributive property.
4. 48 ÷ 3 Solve the following problems using the area model. Support the area model with long division or the distributive property.
5. 49 ÷ 3
6. 56 ÷ 4
7. 58 ÷ 4
8. 66 ÷ 5
9. 79 ÷ 3
10. Seventy-three students are divided into groups of 6 students each. How many groups of 6 students are there? How many students will not be in a group of 6?
NYS Math Module 3 Grade 4 Lesson 21 Homework
2. Solve 79 ÷ 3 using an area model. Use long division and the distributive property to record your work.
3. Paulina solved the following division problem by drawing an area model.
a. What division problem did she solve?
b. Show how Paulina’s model can be represented using the distributive property. 10. Ninety-seven lunch trays were placed equally in 4 stacks. How many lunch trays were in each stack? How many lunch trays will be leftover?
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 534 | 2,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2017-17 | longest | en | 0.895971 |
http://www.numbersaplenty.com/111444444 | 1,571,212,914,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666467.20/warc/CC-MAIN-20191016063833-20191016091333-00526.warc.gz | 295,381,197 | 4,044 | Search a number
111444444 = 2233371672
BaseRepresentation
bin1101010010010…
…00000111011100
321202200222001000
412221020013130
5212012210234
615020350300
72522155626
oct651100734
9252628030
10111444444
11579a8951
12313a5390
131a11c977
1410b2dc16
159bb5899
hex6a481dc
111444444 has 72 divisors (see below), whose sum is σ = 298526480. Its totient is φ = 35927712.
The previous prime is 111444439. The next prime is 111444449. The reversal of 111444444 is 444444111.
Adding to 111444444 its reverse (444444111), we get a palindrome (555888555).
It is an interprime number because it is at equal distance from previous prime (111444439) and next prime (111444449).
It is a Harshad number since it is a multiple of its sum of digits (27).
It is a nude number because it is divisible by every one of its digits.
It is a plaindrome in base 10.
It is a zygodrome in base 10.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (111444449) by changing a digit.
It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 667249 + ... + 667415.
Almost surely, 2111444444 is an apocalyptic number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 111444444, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (149263240).
111444444 is an abundant number, since it is smaller than the sum of its proper divisors (187082036).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
111444444 is a wasteful number, since it uses less digits than its factorization.
111444444 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 384 (or 209 counting only the distinct ones).
The product of its digits is 4096, while the sum is 27.
The square root of 111444444 is about 10556.7250603584. The cubic root of 111444444 is about 481.2301260360.
The spelling of 111444444 in words is "one hundred eleven million, four hundred forty-four thousand, four hundred forty-four". | 611 | 2,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-43 | latest | en | 0.875651 |
https://math.stackexchange.com/questions/760320/computing-the-values-of-w-alphan-int-0-pi-x-alpha-1-sinx2n | 1,563,449,351,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525627.38/warc/CC-MAIN-20190718104512-20190718130512-00372.warc.gz | 473,547,617 | 35,847 | Computing the values of $W_\alpha(n):=\int_0^\pi x^{\alpha-1}\sin(x)^{2n}$
Define the following integral as $$W_\alpha(n):=\int_0^\pi x^{\alpha-1}\sin(x)^{n}\,,\quad V_\alpha(n):=\int_0^\pi x^{\alpha-1}\cos(x)^{n}$$ where $n \in\mathbb{N}$. Now in the base case $W_1(n)=W(n)$ this integral simplifies into the well know Wallis integrals, and $V(n)=W(n)\,\forall\,n$. Which has been shown to be
$$\color{black}{W(n) = 2\int_0^{\pi/2}\sin(x)^n\mathrm{d}x = \left\{ \begin{array}{ll} \frac{\pi}{2^{2p}} \binom{2p}{p} & \text{if} \ \ n = 2p \\ \frac{2^{2p+1}}{p+1} \big/ \binom{2p+1}{n} & \text{if} \ \ n = 2p + 1 \end{array} \right.}$$
Through simmilar means one can also show that $V_2(2n)=W_2(2n)=\pi/2 W(2n)$. By using $$\color{black}{ \int_0^\pi x R(\sin x,\cos^2x)\,\mathrm{d}x=\frac{\pi}{2}\int_0^\pi R(\sin x,\cos^2x)\,\mathrm{d}x }$$
This gives directly that $$W_2(2n) = \int_0^{\pi} x \sin^{2n}(x)\,\mathrm{d}x = \frac{\pi}{2} \int_0^{\pi} \sin^{2n}(x)\,\mathrm{d}x = \pi \int_0^{\pi/2} \sin^{2n}(x)\,\mathrm{d}x = \frac{\pi}{2}W(2n)$$ as wanted, and exactly the same can be done for $V(n)$. My question is:
• Are there any more cases one can solve? When are they alike? Is there a general formula for $W_\alpha(n)$, $V_\alpha(n)$
• Note that $\sin^{2n}x$ and $\cos^{2n}x$ can be written as a finite sum of terms of type $\sin k x$, $\cos k x$ with $k=0,\ldots, 2n$. The remaining integrals can be expressed in terms of the hypergeometric function $_1F_2$. For example: $$\int_0^{\pi}x^{\alpha-1}\cos kx\,dx=\frac{\pi^{\alpha}}{\alpha}{}_1F_2\left[ \begin{array}{c} \alpha/2 \\ 1/2,1+\alpha/2 \end{array};-\frac{\pi^2 k^2}{4}\right]$$ So a general formula exist for both $W_{\alpha}(n)$, $V_{\alpha}(n)$. – Start wearing purple Apr 19 '14 at 12:14
• @N3buchadnezzar I am also started to look into these... I was doing this by using a recurrent relation. Did you figured it out already? What if you have something like [ \int_{0}^\pi x^{\alpha}cos^m(x)sin^n(x) dx ]? This is something I am trying to figure out... Let me know if you have already done it. – user209663 May 21 at 20:52
• @N3buchadnezzar Actually the formula turned out pretty nice after you convert everything into exponential!! – user209663 May 25 at 23:16 | 881 | 2,232 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-30 | latest | en | 0.708056 |
https://tweleve.org/quest-golden-eagle/26725-reverse-engineering-solution-7.html | 1,632,230,933,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057225.38/warc/CC-MAIN-20210921131252-20210921161252-00469.warc.gz | 608,750,621 | 13,532 | # Thread: Reverse Engineering the Solution
1. Lobster,
I commend you on your notes. They will be helpful for the golden eagle hunt.
My worry is that Ron will not be able to sell the golden eagle and the hunt will end early. The reality is that the price of gold is at a five year low and pretty much at the prices that Ron bought it. He will be very lucky to get \$2 mllion for it. Far from the current asking price.
2. Today's imponderable:
"Charlie’s face burned. He had seen enough of women with
plung- ing bodices, heard enough of the coarse laughter, and
seen enough whis- key poured into jelly jars to know what his
mother meant. "
Is whis-key trustable or non-trustable? Normally I'd say no (C5 defines it as "dat poison", which certainly doesn't sound optimistic). However, in this case it is hyphenated with a -key, making it sound like a key hint.
So what's the tie-breaker? Is something that is by default untrustable made trustable by a hyphen? Or is a hyphen overridden by the fact that in the end, it's just a load of poison?
Haven't been able to find those instructions in the manual.
3. I'm visiting the forum again after working on a dozen or so paths for solving Chapter 5; obviously, none successful at this time. My approach varies significantly from this thread's discussion about riddles, word play and clues pointing to other clues across chapters. I realize the four chapter solutions published to date were entirely dependent on Facebook clues. But, I'm curious if anyone is pursuing cipher-based methods, like those published in the companion book or used to solve end-of-chapter questions, to solve a chapter.
Regards,
California Red
4. Good Twelever Gold
Join Date
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This message is only about an observation I noticed, that shouldn't go ignored. Here is the exact quote from Ron:
May 30, 2014 ·
Please note that there is at least one person with at least the first 5 numbers correct. The solution may be close at hand!
Ron
For the participants in C4, you may remember this now. I ask the obvious questions now...how do you get only 5 numbers in C4 and not ten?
I think the answer was in the chapter, and we settled for the second solution with the questions. It explains the frustration, and also the repeated clues in C5. We have not got the message yet on how to do this without added clues.
Does anyone have a solution with CHATB+ outside the official solution?
5. Junior Twelever +1 Bronze
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Iowa
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Originally Posted by gizmo2337
This message is only about an observation I noticed, that shouldn't go ignored. Here is the exact quote from Ron:
For the participants in C4, you may remember this now. I ask the obvious questions now...how do you get only 5 numbers in C4 and not ten?
I think the answer was in the chapter, and we settled for the second solution with the questions. It explains the frustration, and also the repeated clues in C5. We have not got the message yet on how to do this without added clues.
Does anyone have a solution with CHATB+ outside the official solution?
I wonder if someone made a small error(in the 5-10 key range) when extracting the solution from the chapter question, it seems so easy to do. Ron's comment might get one person back on track and throw others off.
My question is when Ron says " the answer they seek is partially spelled" does that mean a word is formed at the beginning of the solution?
6. When I saw this clue in May, I didn't pay attention to the 'close at hand' part. I posted earlier about how if you overlay the chapter picture on to the questions page the shadow of the man connects Q14 to Q8 and I believe the kryptos clue was to point us to the shadow ( between subtle shading and the absence of light). Now notice if you overlay the word 'hand' (the battered one) on page 1 on to the questions page, it overlays the sixth letter in the solution Q14... Hindsight is great eh?
7. Good Twelever Gold
Join Date
Jan 2012
Location
Saanichton
Posts
536
For C4, it sure looks like there was another solution, and Ron reluctantly settled on plan B. I don't think you get past the first letter without getting all ten correct. It would be tough to do I'd think?
In any event, my own take on the "part" or "partially" clue. For C4Q3, the answer to the question is "COLORado". This matches up with C4Q14, where the answer is "Turquoise". The question 3(C) just happens to be the first key, and the matching questions has the rest. C4Q8 and C4Q15 also match this way. "Ditat Deus" means God Enriches, which is related to Q15, the 5x5 book that is the lord's prayer. Note that 5x5=25 is a clue for C5. C4Q16 is your USS Minnesota BB22, because not only do the questions add up to 22, B=2 is also part of the cipher. Bottom line: The non-facebook clue(s) was in the question answers.
8. Dawnplayer,
That is an excellent question. Chapter 4 had a solution starting with the word "chat." Are we possibly going to find this in chapter 5? The W in the word "what" was changed to C, but not by a cipher and this chapter is pointing at a definite cipher of some kind. Easy enough to snare is now ever so easy to snare. To snare is to trap. I have been going back to chapter 4 to see a similiarity with this clue.
9. Originally Posted by gizmo2337
For the participants in C4, you may remember this now. I ask the obvious questions now...how do you get only 5 numbers in C4 and not ten?
Yet another herring. I had the first 10 keys before this clue came out and all this clue did was set me back. Made it sound like alpha is of size 5 when it wasn't.
10. Originally Posted by kazoo
When I saw this clue in May, I didn't pay attention to the 'close at hand' part. I posted earlier about how if you overlay the chapter picture on to the questions page the shadow of the man connects Q14 to Q8 and I believe the kryptos clue was to point us to the shadow ( between subtle shading and the absence of light). Now notice if you overlay the word 'hand' (the battered one) on page 1 on to the questions page, it overlays the sixth letter in the solution Q14... Hindsight is great eh?
Really... Nobody wants to add to this? | 1,506 | 6,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-39 | latest | en | 0.964997 |
https://studytiger.com/modern-electronic/ | 1,542,223,746,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742263.28/warc/CC-MAIN-20181114191308-20181114213308-00379.warc.gz | 753,506,462 | 48,831 | Define Modulation Modulation process of putting information onto a high-frequency carrier for transmission. 2. What Is carrier frequency? In essence, then the transmission takes place at the high frequency (the carrier) which has been modified to "carry' the lower-frequency Information. 3. Describe the two reasons that modulation Is used for communications transmission. The modulated stage accepts two inputs, the carrier and the information (intelligence) signal. 4. List three parameters of a high-frequency carrier that may be varied by a low-frequency intelligence signal. What are the frequency ranges include in the following frequency subdivision: MFC ,HP, VHF, UHF,SF? 300 kHz-MASH 3-MASH VHF= 30-MEZZO UHF=300 MASH-GHZ SF= 3-30 GHZ 9. Convert the following powers to their db equivalents: (d) p=OWE (30 db O (b) p- 0. 001 w (0 db) (c) p=o. 0001 w (-10 db) (d) p=owe (-16 db) 15. Define electrical noise, and explain why it is so troublesome to a communications receiver. Electrical noise may be define as any undesired voltages or currents that ultimately end up appearing in the receiver output. To the listener this electrical noise often manifests Itself as static. 6.
Explain the deference between external and internal noise. External noise in a received radio signal that has been Introduced by the transmitting medium and the Internal noise In a radio signal that has been Introduced by the receiver. 17. List and briefly explain the various types of external noise. Human-Made Noise It is often produced by spark-producing mechanisms such as engine ignition systems,fluorescent lights, and accumulators in electric motors. Atmospheric Noise Is caused by naturally occurring disturbances in the earth's atmosphere, with lightning discharges being the most prominent intrusions. 3. Calculate the SIN ratio for a receiver output of v signal and 0. V noise both as a ratio and in decibel form. (69. 44, 18. Db) 27. A two-stage amplifier has a 3-db bandwidth of kHz determined by an LLC circuit at its input and operates at ICC. The first stage has Peg=db and NFG=2. Db. The second stage has Peg=40th and NFG=6. Db. The out-put Is driving a load of ohms. In testing this system, the noise off 100-k ohms resistor Is applied to Its Input. Calculate the Input and output noise voltage and power and the system noise figure. ( AM 0-1 owe,3. db) 32.
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Calculate the minimum signal power needed for good reception for the receiver (5. XX-IOW) 37. Define information theory. Concerned with optimization of transmitted information. 38. What is Hartley law? Explain its significance. Information that can be transmitted is proportional to the product of the bandwidth times the time of transmission. 40. What is the seventh harmonic of kHz? (kHz's) 49. Define importance and describe its use. Balanced condition between the inductive and capacitive reactant of a circuit. 50. Calculate an inductor's Q at mezzo.
It has an inductance of 6 mm and a series resistance of 1 . K. Determine its dissipation. ( 3. Xx, 0. 318x10-3) 55. Define the quality factor (Q) of an LLC bandannas filter. Explain how it relates to the "selectivity' of the filter. Describe the major limiting value on the Q of a filter. As Q increases, the filter becomes more selective: that is, a smaller passed (narrower bandwidth) is allowed. A major limiting factor in the highest attainable Q is the resistance factor. 65. Draw schematics for Hartley and Solicits oscillators. Briefly explain their operation and differences. | 855 | 3,617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-47 | latest | en | 0.924719 |
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# Quantum mechanics - PowerPoint PPT Presentation
Quantum mechanics. Wave Properties of Matter and Quantum Mechanics I. 5.1 X-Ray Scattering 5.2 De Broglie Waves 5.3 Electron Scattering 5.4 Wave Motion 5.5 Waves or Particles? 5.6 Uncertainty Principle 5.7 Probability, Wave Functions, and the Copenhagen Interpretation
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### Quantum mechanics
• 5.1 X-Ray Scattering
• 5.2 De Broglie Waves
• 5.3 Electron Scattering
• 5.4 Wave Motion
• 5.5 Waves or Particles?
• 5.6 Uncertainty Principle
• 5.7 Probability, Wave Functions, and the Copenhagen Interpretation
• 5.8 Particle in a Box
Louis de Broglie (1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter and radiations, light in particular, it is necessary to introduce the corpuscle concept and the wave concept at the same time.
- Louis de Broglie, 1929
### 5.1: X-Ray Scattering
Max von Laue suggested that if x-rays were a form of electromagnetic radiation, interference effects should be observed.
Crystals act as three-dimensional gratings, scattering the waves and producing observable interference effects.
### Bragg’s Law
• The angle of incidence must equal the angle of reflection of the outgoing wave.
• The difference in path lengths must be an integral number of wavelengths.
Bragg’s Law: nλ = 2d sin θ(n = integer)
William Lawrence Bragg interpreted the x-ray scattering as the reflection of the incident x-ray beam from a unique set of planes of atoms within the crystal.
There are two conditions for constructive interference of the scattered x rays:
### The Bragg Spectrometer
A Bragg spectrometer scatters x rays from crystals. The intensity of the diffracted beam is determined as a function of scattering angle by rotating the crystal and the detector.
When a beam of x rays passes through a powdered crystal, the dots become a series of rings.
In 1925, Davisson and Germer experimentally observed that electrons were diffracted (much like x-rays) in nickel crystals.
George P. Thomson (1892–1975), son of J. J. Thomson, reported seeing electron diffraction in transmission experiments on celluloid, gold, aluminum, and platinum. A randomly oriented polycrystalline sample of SnO2 produces rings.
If a light-wave could also act like a particle, why shouldn’t matter-particles also act like waves?
• In his thesis in 1923, Prince Louis V. de Broglie suggested that mass particles should have wave properties similar to electromagnetic radiation.
• The energy can be written as:
• hn = pc = pln
• Thus the wavelength of a matter wave is called the de Broglie wavelength:
Louis V. de Broglie(1892-1987)
• One of Bohr’s assumptions in his hydrogen atom model was that the angular momentum of the electron in a stationary state is nħ.
• This turns out to be equivalent to saying that the electron’s orbit consists of an integral number of electron de Brogliewavelengths:
electron de Broglie wavelength
Circumference
Multiplying by p/2p, we find the angular momentum:
• De Broglie matter waves should be described in the same manner as light waves. The matter wave should be a solution to a wave equation like the one for electromagnetic waves:
• with a solution:
• Define the wave number k and the angular frequency w as usual:
It will actually be different, but, in some cases, the solutions are the same.
Y(x,t) = A exp[i(kx – wt – q)]
and
C. Jönsson of Tübingen, Germany, succeeded in 1961 in showing double-slit interference effects for electrons by constructing very narrow slits and using relatively large distances between the slits and the observation screen.
This experiment demonstrated that precisely the same behavior occurs for both light (waves) and electrons (particles).
Electron Double-Slit Experiment
Wave-particle-duality solution showing double-slit interference effects for
• It’s somewhat disturbing that everything is both a particle and a wave.
• The wave-particle duality is a little less disturbing if we think in terms of:
• Bohr’s Principle of Complementarity: It’s not possible to describe physical observables simultaneously in terms of both particles and waves.
• When we’re making a measurement, use the particle description, but when we’re not, use the wave description.
• When we’re looking, fundamental quantities are particles; when we’re not, they’re waves.
5.5: Waves or Particles? showing double-slit interference effects for
• Dimming the light in Young’s two-slit experiment results in single photons at the screen. Since photons are particles, each can only go through one slit, so, at such low intensities, their distribution should become the single-slit pattern.
Each photon actually goes through both slits!
One-slit pattern showing double-slit interference effects for
Two-slit pattern
Can you tell which slit the photon went through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
At low intensities, Young’s two-slit experiment shows that light propagates as a wave and is detected as a particle.
Which slit does the electron go through? showing double-slit interference effects for
• Shine light on the double slit and observe with a microscope. After the electron passes through one of the slits, light bounces off it; observing the reflected light, we determine which slit the electron went through.
• The photon momentum is:
• The electron momentum is:
Need lph < d to distinguish the slits.
Diffraction is significant only when the aperture is ~ the wavelength of the wave.
The momentum of the photons used to determine which slit the electron went through is enough to strongly modify the momentum of the electron itself—changing the direction of the electron! The attempt to identify which slit the electron passes through will in itself change the diffraction pattern! Electrons also propagate as waves and are detected as particles.
5.6: Uncertainty Principle: Energy Uncertainty showing double-slit interference effects for
• The energy uncertainty of a wave packet is:
• Combined with the angular frequency relation we derived earlier:
• Energy-Time Uncertainty Principle: .
Momentum Uncertainty Principle showing double-slit interference effects for
• The same mathematics relates x and k: Dk Dx≥ ½
• So it’s also impossible to measure simultaneously the precise values of k and x for a wave.
• Now the momentum can be written in terms of k:
• So the uncertainty in momentum is:
• But multiplying Dk Dx≥ ½ by ħ:
• And we have Heisenberg’s Uncertainty Principle:
How to think about Uncertainty showing double-slit interference effects for
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at the speedometer you got lost.
Kinetic Energy Minimum showing double-slit interference effects for
• Since we’re always uncertain as to the exact position, , of a particle, for example, an electron somewhere inside an atom, the particle can’t have zero kinetic energy:
The average of a positive quantity must always exceed its uncertainty:
so:
5.7: Probability, Wave Functions, and the Copenhagen Interpretation
• Okay, if particles are also waves, what’s waving? Probability
• The wave function determines the likelihood (or probability) of finding a particle at a particular position in space at a given time:
The probability of the particle being between x1 and x2 is given by:
The total probability of finding the particle is 1. Forcing this condition on the wave function is called normalization.
5.8: Particle in a Box Interpretation
• A particle (wave) of mass m is in a one-dimensional box of width ℓ.
• The box puts boundary conditions on the wave. The wave function must be zero at the walls of the box and on the outside.
• In order for the probability to vanish at the walls, we must have an integral number of half wavelengths in the box:
• The energy:
• The possible wavelengths are quantized and hence so are the energies:
Probability of the particle vs. position Interpretation
• Note that E0 = 0 is not a possible energy level.
• The concept of energy levels, as first discussed in the Bohr model, has surfaced in a natural way by using waves.
• The probability of observing the particle between x and x + dx in each state is
Quantum Mechanics II Interpretation
• 6.1 The Schrödinger Wave Equation
• 6.2 Expectation Values
• 6.3 Infinite Square-Well Potential
• 6.4 Finite Square-Well Potential
• 6.5 Three-Dimensional Infinite-Potential Well
• 6.6 Simple Harmonic Oscillator
• 6.7 Barriers and Tunneling
Erwin Schrödinger (1887-1961)
A careful analysis of the process of observation in atomic physics has shown that the subatomic particles have no meaning as isolated entities, but can only be understood as interconnections between the preparation of an experiment and the subsequent measurement.
- Erwin Schrödinger
Opinions on quantum mechanics Interpretation
I think it is safe to say that no one understands quantum mechanics. Do not keep saying to yourself, if you can possibly avoid it, “But how can it be like that?” because you will get “down the drain” into a blind alley from which nobody has yet escaped. Nobody knows how it can be like that.
- Richard Feynman
Those who are not shocked when they first come across quantum mechanics cannot possibly have understood it.
- Niels Bohr
Richard Feynman (1918-1988)
### 6.1: The Schrödinger Wave Equation Interpretation
where V = V(x,t)
The Schrödinger wave equation in its time-dependent form for a particle of energy E moving in a potential V in one dimension is:
where i is the square root of -1.
The Schrodinger Equation is THE fundamental equation of Quantum Mechanics.
### General Solution of the Schrödinger Wave Equation when InterpretationV = 0
Try this solution:
This works as long as:
which says that the total energy is the kinetic energy.
### General Solution of the Schrödinger Wave Equation when InterpretationV = 0
In free space (with V = 0), the general form of the wave function is
which also describes a wave moving in the x direction. In general the amplitude may also be complex.
The wave function is also not restricted to being real. Notice that this function is complex.
Only the physically measurable quantities must be real. These include the probability, momentum and energy.
### Normalization and Probability Interpretation
The probability P(x) dx of a particle being between x and x + dx is given in the equation
The probability of the particle being between x1 and x2 is given by
The wave function must also be normalized so that the probability of the particle being somewhere on the x axis is 1.
Properties of Valid Wave Functions Interpretation
• Conditions on the wave function:
• 1. In order to avoid infinite probabilities, the wave function must be finite everywhere.
• 2. The wave function must be single valued.
• 3. The wave function must be twice differentiable. This means that it and its derivative must be continuous. (An exception to this rule occurs when V is infinite.)
• 4. In order to normalize a wave function, it must approach zero as x approaches infinity.
• Solutions that do not satisfy these properties do not generally correspond to physically realizable circumstances.
Time-Independent Schrödinger Wave Equation Interpretation
• The potential in many cases will not depend explicitly on time.
• The dependence on time and position can then be separated in the Schrödinger wave equation. Let:
• which yields:
• Now divide by the wave function y(x) f(t):
The left side depends only on t, and the right side depends only on x. So each side must be equal to a constant. The time dependent side is:
Time-Independent Schrödinger Wave Equation Interpretation
We integrate both sides and find:
where C is an integration constant that we may choose to be 0.
Therefore:
But recall our solution for the free particle:
In which f(t) = e -iw t, so: w = B / ħ or B = ħw, which means that: B = E !
So multiplying by y(x), the spatial Schrödinger equation becomes:
Time-Independent Schrödinger Wave Equation Interpretation
This equation is known as the time-independent Schrödinger wave equation, and it is as fundamental an equation in quantum mechanics as the time-dependent Schrodinger equation.
So often physicists write simply:
where:
is an operator.
Stationary States Interpretation
• The wave function can be written as:
• The probability density becomes:
• The probability distribution is constant in time.
• This is a standing wave phenomenon and is called a stationary state.
6.2: Expectation Values Interpretation
• In quantum mechanics, we’ll compute expectation values. The expectation value, , is the weighted average of a given quantity. In general, the expected value of x is:
If there are an infinite number of possibilities, and x is continuous:
Quantum-mechanically:
And the expectation of some function of x, g(x):
Momentum Operator Interpretation
• To find the expectation value of p, we first need to represent p in terms of x and t. Consider the derivative of the wave function of a free particle with respect to x:
• With k = p / ħwe have
• This yields
• This suggests we define the momentum operator as .
• The expectation value of the momentum is
Position and Energy Operators Interpretation
The position x is its own operator. Done.
Energy operator: The time derivative of the free-particle wave function is:
Substituting w = E / ħ yields
The energy operator is:
The expectation value of the energy is:
Deriving the Schrodinger Equation using operators Interpretation
The energy is:
Substituting operators:
E:
K+V:
Substituting:
6.3: Infinite Square-Well Potential Interpretation
• The simplest such system is that of a particle trapped in a box with infinitely hard walls thatthe particle cannot penetrate. This potential is called an infinite square well and is given by:
• Clearly the wave function must be zero where the potential is infinite.
• Where the potential is zero (inside the box), the time-independent Schrödinger wave equation becomes:
• The general solution is:
x
0
L
where
Quantization Interpretation
• Boundary conditions of the potential dictate that the wave function must be zero at x = 0and x = L. This yields valid solutions for integer values of n such that kL = np.
• The wave function is:
• We normalize the wave function:
• The normalized wave function becomes:
• These functions are identical to those obtained for a vibrating string with fixed ends.
x
0
L
½ - ½ cos(2npx/L)
Quantized Energy Interpretation
• The quantized wave number now becomes:
• Solving for the energy yields:
• Note that the energy depends on integer values of n. Hence the energy is quantized and nonzero.
• The special case of n = 1 is called the ground state.
6.4: Finite Square-Well Potential Interpretation
• The finite square-well potential is
The Schrödinger equation outside the finite well in regions I and III is:
Letting:
yields
Considering that the wave function must be zero at infinity, the solutions for this equation are
Finite Square-Well Solution Interpretation
• Inside the square well, where the potential V is zero, the wave equation becomes where
• The solution here is:
• The boundary conditions require that:
• so the wave function is smooth where the regions meet.
• Note that the wave function is nonzero outside of the box.
Penetration Depth Interpretation
• The penetration depth is the distance outside the potential well where the probability significantly decreases. It is given by
• The penetration distance that violates classical physics is proportional to Planck’s constant.
6.6: Simple Harmonic Oscillator Interpretation
• Simple harmonic oscillators describe many physical situations: springs, diatomic molecules and atomic lattices.
Consider the Taylor expansion of a potential function:
Simple Harmonic Oscillator Interpretation
Consider the second-order term of the Taylor expansion of a potential function:
Substituting this into Schrödinger’s equation:
Let and which yields:
The wave function solutions are where InterpretationHn(x) are Hermite polynomials of order n.
The Parabolic Potential Well
The Parabolic Potential Well Interpretation
• Classically, the probability of finding the mass is greatest at the ends of motion and smallest at the center.
• Contrary to the classical one, the largest probability for this lowest energy state is for the particle to be at the center.
Analysis of the Parabolic Potential Well Interpretation
As the quantum number increases, however, the solution approaches the classical result.
The Parabolic Potential Well Interpretation
• The energy levels are given by:
The zero point energy is called the Heisenberg limit:
6.7: Barriers and Tunneling Interpretation
• Consider a particle of energy E approaching a potential barrier of height V0,and the potential everywhere else is zero.
• First consider the case of the energy greater than the potential barrier.
• In regions I and III the wave numbers are:
• In the barrier region we have
Reflection and Transmission Interpretation
• The wave function will consist of an incident wave, a reflected wave, and a transmitted wave.
• The potentials and the Schrödinger wave equation for the three regions are as follows:
• The corresponding solutions are:
• As the wave moves from left to right, we can simplify the wave functions to:
Probability of Reflection and Transmission Interpretation
• The probability of the particles being reflected R or transmitted T is:
• Because the particles must be either reflected or transmitted we have: R + T = 1.
• By applying the boundary conditions x→ ±∞, x = 0, and x = L, we arrive at the transmission probability:
• Note that the transmission probability can be 1.
Tunneling Interpretation
• Now we consider the situation where classically the particle doesn’t have enough energy to surmount the potential barrier, E < V0.
The quantum mechanical result is one of the most remarkable features of modern physics. There is a finite probability that the particle can penetrate the barrier and even emerge on the other side!
The wave function in region II becomes:
The transmission probability that describes the phenomenon of tunneling is:
Tunneling wave function Interpretation
This violation of classical physics is allowed by the uncertainty principle. The particle can violate classical physics by DE for a short time, Dt ~ ħ / DE.
Analogy with Wave Optics Interpretation
• If light passing through a glass prism reflects from an internal surface with an angle greater than the critical angle, total internal reflection occurs. However, the electromagnetic field is not exactly zero just outside the prism. If we bring another prism very close to the first one, experiments show that the electromagnetic wave (light) appears in the second prism The situation is analogous to the tunneling described here. This effect was observed by Newton and can be demonstrated with two prisms and a laser. The intensity of the second light beam decreases exponentially as the distance between the two prisms increases.
Alpha-Particle Decay Interpretation
• The phenomenon of tunneling explains alpha-particle decay of heavy, radioactive nuclei.
• Inside the nucleus, an alpha particle feels the strong, short-range attractive nuclear force as well as the repulsive Coulomb force.
• The nuclear force dominates inside the nuclear radius where the potential is ~ a square well.
• The Coulomb force dominates outside the nuclear radius.
• The potential barrier at the nuclear radius is several times greater than the energy of an alpha particle.
• In quantum mechanics, however, the alpha particle can tunnelthrough the barrier. This is observed as radioactive decay.
6.5: Three-Dimensional Infinite-Potential Well Interpretation
• The wave function must be a function of all three spatial coordinates.
• Now consider momentum as an operator acting on the wave function. In this case, the operator must act twice on each dimension. Given:
So the three-dimensional Schrödinger wave equation is
The 3D infinite potential well Interpretation
It’s easy to show that:
where:
and:
When the box is a cube:
Try 10, 4, 3 and 8, 6, 5
Note that more than one wave function can have the same energy.
Degeneracy Interpretation
• The Schrödinger wave equation in three dimensions introduces three quantum numbers that quantize the energy. And the same energy can be obtained by different sets of quantum numbers.
• A quantum state is called degenerate when there is more than one wave function for a given energy.
• Degeneracy results from particular properties of the potential energy function that describes the system. A perturbation of the potential energy can remove the degeneracy. | 4,695 | 21,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-51 | latest | en | 0.820615 |
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# Min+Shear+Reinf - 3 Minimum Area Requirements for Shear...
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Unformatted text preview: 3. Minimum Area Requirements for Shear Reinforcement AC1 Code Section 11.4.6. - This criteria must be satisfied in "normal" beams whenever V“ /¢ > V204) 5mm m ova/Z bws (1143) fyr - Value used for 0.75 J}: must be 2 50 psi — For consistency, I like to solve this for the Spacing to satisfy the minimum area re uirement, i.e.: S < Atfy, ‘ bw(0.75)\[}7: 4. Maximum Spacing Criteria for Shear Reinforcement AC1 Code Section 11.4.5 — Again, this requirement is to be satisfied whenever V" /¢ > 1/2( V6) for Vs(req'd.)=(Vu/¢- VC) 5 4 J17; (bwd), s s an and \$24" for Vs(req'd.)= (14, /¢- 19> 4 J? (b... d), s Sci/4 and 512" — Note that Code Section 11.4.7.9 states that if, V.(req'd.)=(n/¢- Va > 8 if? (bid) the section area must be increased. 5. Notes on Design Shear Envelope at Compression Supports i. AC1 Code Section 11.1.3 allows ”in most cases” for the end region of a beam to be designed for the "design shear" occurring at a distance "d" from the face of the support. Exceptions are for: (a) concentrated loads within "d" of the support face, (b) a "noncompression" type of support, i.e., a beam (girder) supporting a beam, and (c) loads applied below mid-depth of the beam. ii. Place first stirrup at 5/2 from face of support. ...
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Jill Tulane University ‘16, Course Hero Intern | 695 | 2,512 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-22 | latest | en | 0.839607 |
https://www.physicsforums.com/threads/what-is-limit-below-this.818905/ | 1,511,542,770,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808260.61/warc/CC-MAIN-20171124161303-20171124181303-00440.warc.gz | 839,241,547 | 20,345 | # What is limit below this?
1. Jun 13, 2015
### Md. Abde Mannaf
We know,
1/0.1=10;
1/0.01=100;
1/0.001=1000; 1/0.0001=10000;
.
.
.
1/0.00000000001=100000000000
.
.
.
1/0=infinity
Why is it not correct?
2. Jun 13, 2015
### Staff: Mentor
Infinity is not a real number.
Also consider:
1/(-0.1)=-10;
1/(-0.01)=-100;
...
1/0 = -infinity?
1/(-0.1)=-10;
1/(0.01)=100;
1/(-0.001)=-1000;
1/(0.0001)=10000;
1/0 = ???
3. Jun 13, 2015
### disregardthat
There's a reason why mathematicians say 1/0 is undefined. It just doesn't fit in with the general framework. You are free to define it to be infinity, or whatever you want, but beware that you cannot use the standard rules of arithmetic to manipulate fractions containing 0 as a denominator. You will quickly run into consistencies.
4. Jun 13, 2015
### DaveC426913
Here's why you will run into trouble.
The operation of division is defined as the converse of multiplication.
Note here, how the first thing they do to define division is to start with multiplication and invert it:
https://en.wikipedia.org/wiki/Division_(mathematics)
So, while it is true that
10 * 0.1 = 1
100 * 0.01 = 1
1000 * .001 = 1
it is not true that
infinity * 0 = 1.
Thus it's converse (1/0=infinity) is also not true.
That bears reiteration: School taught you the shortcut for division, but once you start pushing at the leaky corners of numbers, you've got to use more formal maths.
Last edited: Jun 13, 2015
5. Jun 13, 2015
### Md. Abde Mannaf
if i devided 10 apple between no one then every one get infinity? thats not correct. so you devide something by zero does not make infinite. its just undefined.
you can get infinite if 1/ x with x tens to zero. but exactly zero will give you undefined sol.
6. Jun 13, 2015
### DaveC426913
No, you will just get a very large number.
7. Jun 13, 2015
### Md. Abde Mannaf
very very large number is not infinity?
8. Jun 14, 2015
### Svein
No. Since nothing (in this sense) is larger than infinity, a very large number is not infinity because:
• You give me an extremely large number (call it P). I answer "but P+1 is larger, so P cannot be infinity"
9. Jun 14, 2015
### micromass
Staff Emeritus
Yes in the sense that the limit of $1/x$ when positive $x$ tends to $0$ is infinity. What you are describing in your OP is exactly what limits do, they don't tell us what $1/0$ is exactly, but they do tell how small numbers $\varepsilon>0$ behave when we do the division $1/\varepsilon$. In this case, if $\varepsilon$ get very close to $0$, then $1/\varepsilon$ will get bigger and bigger, and thus will be infinity in the limit.
However, you can only do this when talking about limits. Doing arithmetic with infinity is way more technical when not doing limits.
10. Jun 14, 2015
Talking about apples. If you divide 10 apples between zero people, you didn't even divide them. You did nothing, therefore your action makes no sense and it's not defined. Both infinity and nothing are very interesting/complicated terms that probably no one can understand.
@micromass : Talking about limes, I always thought of it as something not so... clean. For example, if you have 1 + 1/2 + 1/4 + 1/8 + ... you cannot just simply say that it all combined gives 2. It will never give 2 whatever you do. I actually had an argument with my math teacher in school about this. I mean, you can say "Yeah, this sum is leaning towards 2", but in fact it will never be 2.
Last edited: Jun 14, 2015
11. Jun 14, 2015
### jbriggs444
In the context of limits, infinity is not a number. It is more of a boundary. No number is larger than infinity in this sense. No number is even equal to infinity.
When one sees "x-> oo" used in the bounds of a limit, it is an abuse of notation that should be read "as x increases without bound".
When one sees "oo" used as the value of a limit, it is an abuse of notation that should be read as a statement that the limit does not exist and that it does not exist in a particular way.
One can go a step further and think of numbers that are larger and larger as being closer and closer to "infinity". This involves modifying our usual notion of "closeness" where we decide how close two numbers are by looking at the absolute value of their difference. [Since infinity is not a number, that difference would be undefined]. This process amounts to defining a "topology" on an extended set that includes two new members: +oo and -oo.
Defining this topology effectively gives you two new positions on an extended number line. The old number line had no end points. The new one has a new right hand endpoint and a new left hand endpoint. This process can be called "the two point compactification of the reals". Doing this just defines the new positions. It does not necessarily define any arithmetic operations on those positions.
With this two point compactification in hand, one can go back and re-interpret the limit notation
When one sees "x -> oo" used in the bounds of limit, it is correct notation that can be read as "as x approaches positive infinity"
When one sees "oo" used as the value of the limit, it is correct notation that indicates that the limit is infinity.
Note that although it is tempting to go ahead and define arithmetic operations on +oo and -oo, it is not possible to do so in a way that preserves all of the axioms of the real numbers as an ordered field. Accordingly, although one can use +oo and -oo as bounds for limits or as the result of a limit, one cannot use them in ordinary arithmetic operations or produce them as the result of an ordinary arithmetic operation.
12. Jun 14, 2015
### Staff: Mentor
"limits" are what we're talking about, not limes, which are citrus fruits.
This sum, 1 + 1/2 + 1/4 + 1/8 + ... is equal to 2. The ... (an ellipsis) indicates that we are talking about the sum of an infinite number of terms, with each term following a certain pattern. It's easy to show that the partial sum Sn, 1 + 1/2 + 1/4 + 1/8 + ... + 1/2n - 1, can be made arbitrarily close to 2. This shows that the limit of the partial sums, $\lim_{n \to \infty} S_n = 2$.
Presumably your teacher won the argument.
What you say would be correct if you were talking about the sum of a finite number of terms, which isn't the case here.
13. Jun 14, 2015
### micromass
Staff Emeritus
Note that the sum $1+\frac{1}{2} + \frac{1}{4} + ...$ is defined as the limit of the finite sums! So the equality is by definition. If you want to use another interpretation of the infinite sum, be my guest, but the mathematicaly community has agreed on this definition.
14. Jun 14, 2015
@Mark44 and @micromass Yes, I am aware of that and I am aware of the definitions regarding geometric sum in this case. I'm just saying that to me it just doesn't seem right to simply say that the sum I was talking about is equal to 2 without specifically stating that you are looking for the limit of that sum. Yes, that is just my interpretation and when I'm re-reading it, I've noticed that I wasn't able to really explain what I meant as English is not my native language. Also, could someone please explain me how to write fractions and stuff like that here in the posts? Zeno's paradox as in Achilles and the tortoise?
15. Jun 14, 2015
### micromass
Staff Emeritus
If you say that the sum can't equal $2$, then you must first provide a definition of the infinite sum.
16. Jun 14, 2015
17. Jun 14, 2015
### Staff: Mentor
There are a couple of sums here.
1. Partial sum:
Sn = 1 + 1/2 + 1/4 + ... + 1/2n - 1, the sum of the first n terms. Another way to write this is
$\sum_{i = 0}^{n - 1} \frac 1 {2^i}$
2. Infinite sum:
S = 1 + 1/2 + 1/4 + ... + 1/2n - 1 + 1/2n + ...
The latter sum is exactly equal to 2. The way this is proved is by showing that the sequence of partial sums, {Sn}, gets arbitrarily close to 2.
When you say it doesn't seem right that the latter sum equals 2, all it means is that you don't quite understand how an infinite series (such as the one you posted) can converge to a number.
LaTeX fractions:
# # \frac{1}{2^3}# # (omit the spaces between the # characters)
18. Jun 14, 2015
### DaveC426913
Of course not. | 2,228 | 8,159 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-47 | longest | en | 0.89985 |
https://mathjokes4mathyfolks.wordpress.com/2013/09/01/you-say-its-your-birthday/ | 1,660,286,865,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571584.72/warc/CC-MAIN-20220812045352-20220812075352-00709.warc.gz | 360,724,511 | 26,695 | ## You Say It’s Your Birthday…
Well, no, actually it’s not my birthday. And it’s not my friend Jacqui’s birthday, either, but she did just celebrate a milestone with us that she wanted to share. Via email, she announced,
I’ve been alive for two billion seconds, a milestone I passed this morning.
This reminded me of a problem from Steve Leinwand’s book, Accessible Mathematics, in which he tells kids his age as a unitless number, then asks them to identify what units he must be using. Along those lines, here are some questions for you.
• How old (in years) is my friend Jacqui?
• What is her date of birth?
• If I tell you that my age is 22,333,444, what units must I be using? Assuming I’m not telling a fib, of course. And what is my age in years and my date of birth?
This reminds me of two math jokes about birthdays.
Statistics show that those who celebrate the most birthdays live longest.
An algebraist remembers that his wife’s birthday is on the (n – 1)st of the month. Unfortunately, he only remembers this when he is reminded on the nth.
Entry filed under: Uncategorized. Tags: , , , .
The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.
## MJ4MF (offline version)
Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature. | 382 | 1,626 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-33 | latest | en | 0.923591 |
https://www.studypug.com/ca/grade6/taxes-discounts-tips-and-more | 1,675,045,285,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499790.41/warc/CC-MAIN-20230130003215-20230130033215-00438.warc.gz | 1,035,439,088 | 35,865 | # Taxes, discounts, tips and more
#### All in One Place
Everything you need for better grades in university, high school and elementary.
#### Learn with Ease
Made in Canada with help for all provincial curriculums, so you can study in confidence.
#### Instant and Unlimited Help
0/3
##### Intros
###### Lessons
1. Introduction to taxes, discounts, tips and more
2. Relating Percents to Taxes
3. Relating Percents to Tips
4. Relating Percents to Discounts
0/5
##### Examples
###### Lessons
1. Relating Percents to Taxes
A pack of potato chips is priced at $5. If the sales tax in the province is 15%, how much should the potato chips be sold for? 1. Billy picked a pair of headphones with a selling price of$250 at an electronic store. However, at the check-out counter, he was charged $280. What is the percentage of the sales tax? 1. Relating Percents to Tips A family of three went to a five-star restaurant for dinner. The bill for the meal was$180. The family put down $300 and left without asking for changes. In other words, what is the percentage of tips they have paid? 1. Relating Percents to Discounts A stationary store is having a garage sale. All the binders priced at$12 are now on sale for $9. What is the discount percentage? 1. A suit was originally priced at$880, but it has been marked 25% off. What is the selling price now?
###### Topic Notes
(Original price) + [(Original price) $\times$ (tax/tip in percent)] = (New price)
(Original price) $\times$ [1+ (tax/tip percent in decimal)] = (New price)
(Original price) - [(Original price) $\times$ (discount in percent)] = (New price)
(Original price) $\times$ (1 - discount percent in decimal) = (New price) | 426 | 1,687 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-06 | latest | en | 0.916867 |
https://dojo.domo.com/discussion/52681/domo-ideas-conference-beast-modes-running-totals | 1,653,292,021,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662556725.76/warc/CC-MAIN-20220523071517-20220523101517-00299.warc.gz | 273,423,364 | 19,602 | # Domo IDEAs Conference - Beast Modes - Running Totals
Indiana π₯·
Greetings! This is another post highlighting a beast mode from my Domo IDEAs conference session. This one covers how to calculate a grand total, running total overall and by month.
Problem:
How do I calculate a running total?
Solution:
We can utilize a window function which will calculate some metrics across the entire dataset (or a subset / partition of it) and return that value for each row.
You'll notice that there is a SUM(SUM(..)) call which may seem odd. Traditional SQL only has a single aggregate function. Because the way Domo works if there is another column being aggregated that aggregation happens after the beast mode. So because a single value is being returned Domo is attempting to aggregate it but doesn't know how to so it'll throw an error if there isn't a second aggregation.
Grand Total:
```-- Author:
-- Created:
-- Description: Using a window function return the grand total for the entire dataset
-- Will return this value for each row.
SUM(SUM(`random_number`)) OVER ()
```
Running Total:
Running totals require the ORDER BY clause to know how to order the data when doing a running total.
```-- Author:
-- Created:
-- Description: Get the running total over the entire dataset. (Not bucketed / partitioned)
SUM(SUM(`random_number`)) OVER (ORDER BY `dt`)
```
Running Total By Month:
This utilizes the `PARTITION BY` function to specify the group in which to perform the running total.
```-- Author:
-- Created:
-- Description: Calculate the monthly running totals.
-- PARTITION BY states how to bucket the information, in this case each month and year.
-- ORDER BY tells the order in performing the running calculation
-- Without the ORDER BY it would just return the total for the entire month on each row
SUM(SUM(`random_number`)) OVER (PARTITION BY YEAR(`dt`), MONTH(`dt`) ORDER BY `dt`)
```
Note:
Window functions are not enabled by default and require this feature be switched on in your instance. Talk with your CSM to get it enabled.
**Did this solve your problem? Accept it as a solution!**
• Indiana π₯·
Here's a link to the video of my session outlining this beast mode: https://www.youtube.com/watch?v=gO8OLpsAk4M&index=6
**Did this solve your problem? Accept it as a solution!**
• π’
@GrantSmith have you run into any issues getting these beast modes to work correctly with the analyzer date filter (i.e. this year graph by week/month etc)?
I'm trying to add a running percent of total line to a chart (running total of x/running total of y) following this and the values are correct as long as I'm not using any of the analyzer's group by options in the date range.
• Indiana π₯·
This isn't compatible with the group by options in the date range because we're defining how we want to do the grouping in the window function with the PARTITION BY section. | 657 | 2,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-21 | latest | en | 0.862862 |
http://www.amazinavenue.com/2011/8/2/2339273/the-remaining-schedule | 1,409,238,874,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500830903.34/warc/CC-MAIN-20140820021350-00291-ip-10-180-136-8.ec2.internal.warc.gz | 238,618,126 | 23,406 | ## The remaining schedule
As uplifting as the sweep of the Reds was, the recent string of suck has deflated my hopes for the WC. I see no chance that the Mets make it. I could go on a rant about this guy's defense, or that guy's relief pitching, or the other guy's starting pitcher, but in the end, the sum of all fears (the sum of all fails) adds up to "does not compute" as far as making the WC. With the season running out of runway, let's have a look at the remaining games in the schedule:
Marlins = We have 9 games left with them. We have proven that we are unable to beat them over the last 4 years. In a best case scenario, we go 5-4 against them.
Braves = Also 9 games left against them. Like the Marlins, we are also unable to beat them (not to mention that they are the better team). For the sake of optimism, I will say 5-4 the rest of the way.
Padres = 6 games with them. We should probably go 4-2.
D-Backs = 3 games with them in Arizona. I'll say 1-2.
Brewers = 3 games. 2-1
Phillies = 6 games. 2-4
Nats = 7 games. See Marlins and Braves above. 3-4
Cubs = 3 games. 2-1
Cards = 3 games 1-2
Reds = 3 games 2-1
If my math is correct, this scenario bring us to a final record of 82-80. This is of course, assuming that there are no further injuries to the main players, and also assumes that players wont stick their heads up their ass. To make the WC, assuming the Braves play at .500 the rest of the way, the Mets need to play 7 games better than them just to catch them. I don't know where we find 7 wins in this schedule. Even throwing away logic and changing predictions to 6-3 vs Marlins and Braves, change Phillies to 3-3, and sweep Cubs and Reds, we still fall short. Also, there is no indication that the Braves will play at .500 the rest of the way. Most likely they will play above .500 assuming everything stays the same from an injury / performance on the field standpoint. So, friends.....it is time to dust off the AAOP and start dreaming about ST.
Update as of August 6th = after losing 5 in a row, and dropping back below .500, the Mets are 9 back with 51 to play. Again, assuming the Braves play .500 from today on, they will finish with 90 wins. For the Mets to reach 90 wins, they need to go 35-16 the rest of the way.
This FanPost was contributed by a member of the community and was not subject to any vetting or approval process. It does not necessarily reflect the opinions, reasoning skills, or attention to grammar and usage rules held by the editors of this site.
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We have our own Community Guidelines at Amazin' Avenue. You should read them. | 865 | 3,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2014-35 | latest | en | 0.944822 |
https://id.scribd.com/document/337725377/Integral-Representation-of-Kelvin-Functions | 1,563,447,631,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525627.38/warc/CC-MAIN-20190718104512-20190718130512-00443.warc.gz | 424,164,396 | 67,212 | Anda di halaman 1dari 7
# Journal of Applied Mathematics and Physics (ZAMP)
## 0044-2275/91/050708-07\$ 1.50 + 0.20
9 1991 Birkh~iuser Verlag, Basel
## Integral representation of Kelvin functions and their
derivatives with respect to the order
By Alexander Apelblat, Department of Chemical Engineering,
Ben Gurion University of the Negev, Beer Sheva, Israel
1. Introduction
In 1890 Lord Kelvin (Sir William Thomson) [1] denoted the real and
imaginary parts of the solution, finite at the origin, of the following
differential equation
xy"+y'--ixy
=0
(1)
## as b e r x and bei x. The meaning of these symbols is: Bessel-real and
Bessel-imaginary parts of the Bessel function of order zero and of the
a r g u m e n t i3/2x. Equation (1) appears in solution of certain problems of
electrical engineering. The functions under consideration were introduced,
at least implicitly, a few years earlier in 1884 by Heaviside [2], who
investigated a similar electrical problem. Whitehead in 1911 [3] denoted the
solutions of the following differential equation
xZy " + x y ' - (v 2 + ix2)y = 0
(2)
as berv x, be% x, ker~ x and keiv x where v is real and non-negative. These
functions are called the Kelvin functions of order v and of the first and
second kind respectively.
The Kelvin functions were investigated because they are involved in
solutions of various engineering problems occurring in the theory of electrical currents, elasticity and in fluid mechanics [4-6].
The basic formulas and properties of different Kelvin functions are
given in the literature [7-10], especially in the McLachlan book [11].
However, with an exception of asymptotic expansions [ 10], the properties of
the Kelvin functions related to operations with respect to their order are
virtually unknown. The same situation exists with regard to the integral
representations of these functions. This gap is fulfilled in this work. The
present study is a continuation of our previous investigation dealing with
the Bessel functions [12, 13], which evidently is a starting point for the
Kelvin functions.
709
## 2. Kelvin functions ber~ x and beiv x
The Kelvin functions of the order v are related to the Bessel functions
in the following way:
berv x + beiv x = e +-i~vJ~(xe
(3)
T-i~/4)
## where v is real and x is real and non-negative.
The integral representation of the Bessel function J~(z) in the Schlaefli
form is [10]
Jr(z)
--/'C
fo
f0
cos(z sin x - v x ) d x
sin T/;V
....
sinh x
dx
(4)
From (3) and (4), after long but rather elementary transformations, the
integral representations of the Kelvin functions are:
berv(xx/~) = _1
7~
sin zcv
7C
f0
--vt-
sinht COS(X
sinh t + ~v) dt
(5)
and
beiv(Xx/~)
7Z
## + sin rcv cos(x sin t - vt) cosh(x sin t)] dt
sin~rrcv f f e -vt- x sinh, sin(x sinh t + roy) dt
(6)
## For a positive integer n or zero they are reduced to:
ber~(xx/~) = ( - 1)~-
## cos(x sin t - nt) cosh(x sin t) dt
(7)
bein(xx/~) - ( - 1)n
## sin(x sin t - nt) sinh(x sin t) dt
(8)
7/
and
By changing the integration variable in (7) and (8), the Kelvin functions of
the zero order can be presented in the form:
ber(xx/~ ) =-re
2 fo 1 cos(xt) cosh(xt) x/~dt_ t~
(9)
710
Alexander Apelblat
ZAMP
and
2 foI sin(xt)
bei(xx//-~) =-re
sinh(xt) ~
dt_
t2
(10)
T h e Kelvin functions bern + 1/2 x a n d bein+ 1/2 x, can consecutively be evaluated in terms o f elementary functions using the recurrence relations [10]:
berv + 1 x + ber~ _ 1 x = -
v,/5(
ber~ x - beiv x)
( 11 a)
bei~+ 1 x + bei~_l x = -
(bei~ x + berv x)
(llb)
and
_
2 -3/4
x/~{eXcos(x+8)+e-Xcos(x-8)}
berl/2(xx//~)
_
beil/2(xx/~)
(12a)
2 -3/4
x/~{eXsin(x+8)+e-Xsin(x-8)}
(12b)
and
_
2 -3/4
b e r - 1/2( x x / - 2 ) , ~ - s 1 6 3
~{eXcos(x+8)-e-Xcos(x-8)
bei_ ~/2(x~//-2) -
(13a)
(13b)
ber* (x) -
berv(x)
Ov
(14a)
bei* (x) -
bei~(x)
~?v
(14b)
## and differentiating (5) with respect to v, we have:
ber*(xx//-2 ) = - r e bei,,(xx/~)
+ -
7~
## - cos ~zv sin(x sin
t - vt) sinh(x
t - vt) cosh(x
sin t)]
sin t)
dt
+ - 1 j ' ~ e - ~t- x sinh t(t sin 7zv -- rc cos 7zv) cos(x sin t + rcv)
7~
dt
(15)
## Integral representation of Kelvin functions
71 l
and
lfo
bei*(xv/2 ) = ~ berv(xv/2 ) + -
7~
+ -1 fo~ e
## sin ltv - 7t cos ~zv) sin(x sin t + ~v) act
= vt - x sinh t(t
7~ 0
(16)
In order to express (15) and (16) in equivalent forms let us introduce:
F(v, y, t) = t ~/2 berv(yx//t)
(17)
## Performing differentiation of Eq. (17) with respect to v we have:
F*(v, y, t) = t v/2 lnw/t berv(yx//t) + t ~/2 ber*(yx/~ )
(18)
and therefore
F*(v, xx//2, 1) = ber*(xx/~ )
(19)
## The Laplace transform of F(v, y, t) is given by: [13, p. 167, (16.42)]
L { t u/2 ber~( y x/~) } = G(v, y, s)
s "~s cos
(20a)
+ 4 ]'
Res>0,
Rev>-i
(20b)
and evidently
L _ I~ i~G(v, Y,
s)} = F*(v, Y, t)
(21)
av
s ~s
In
cos
,,(,)v
4s ~s
sin
(22)
## but [14, p.167, (16.43)]
L{tU/2beiv(yx/Tt)}=s
I(Y) v
~s
sin
[~--52+---~--]
3~v7 Res >0,
Rev>-l
(23)
712
AlexanderApelblat ZAMP
## F*(v, y, t) - --3---~tv/24 bei~(Yx/Ft)+V/2 ln(2)ber~(yv/t)
-- Z - 1{~--~(-~s)VCOSI~ "1--~'~1}
(24)
but
(-~ss)~C~
-1
2
+--4-J=--~s3rCv-]
yw/2(y){coS[~s+
-
sin
3rc(v-]4- 1)1
## and [14, p. 268, (6.1)]
L-'{~S} =- ? -ln t
(26)
## where ? = - ~(1) = 0.57721566...
From the convolution theorem of the Laplace transform and (24) we
have:
## F*(v,y, t)= -3rttv/24 bei~(Yx/~)+ V/2 ln(y) ber~(Yx/~)
Jo u(~ -1)/2[y + ln(t - u)]
x{berv_l(yw/-u)+bei~_l(yw/~)}du,
Rev>0
(27)
## and introducing y = xv/-2 and t = 1, the final result is:
ber, (xx/~) = ln(~22 ) berv(Xx/~) --4-3z~bei~(xx/~)
Xo'
2
## x {ber~_l(Xx/~ ) +beiv_l(xx/~)} du, Rev > 0 (28)
Evaluated in a similar way, the equivalent expression for the second Kelvin
function is:
bei*(xx/~ ) = In ( ~ )
+ -~
(29)
713
## The order derivatives for v = 0, are evaluated using (24):
F*(O,y, 1)= -3re
beiv(yw/t)+4 ln(2)berv(yx/~)
(30)
## but [14, p. 268, (6.8)1
L ,.J'ln s]
1 (7 + 2 In 2 + In t)
(31)
L - ' ~<15 cos
cosh y
cos y
(32)
## and therefore from (30):
ber*(xx/2) = In( ~ )
3re bei(xw/~ )
ber(xw/2 ) --~-
## + - 1 ~1 [7 + 2 In 2 + ln(1 - u)] cosh(xv/~) cos(xw/~ ) du
J0
(33)
and similarly:
bei*(xv/2) = In ( ~ )
3re ber(xw/~ )
bei(xx/~ ) + -~-
## + - 1 f l [7 + 2 In 2 + ln( 1 - u)] sinh(x,Ju) sin(xv/-s ) du
3o
x/~l-u)
(34)
Denoting ? = 2 in C (C = 1.3346682...) and changing the integration variable in (33) and (34) we have:
ber*(xx/~) = in (~22) ber(xx/~ ) - - ~3re bei(x x/~ )
4
+-
f re/2
ln(2C sin O) cosh(xx/-2 cos O)
7C,)o
x cos(x,f2 cos 0) dO
(35)
and
bei*(xx//2) = In
bei(xv/2) + ~ ber(xx/~ )
4
+-
j%r
## ln(2C sin 0) sinh(xx/~ cos 0) sin(xx/~ cos 0) dO
(36)
714
Alexander Apelblat
ZAMP
References
[1] Lord Kelvin (Sir William Thomson), Ether, electricity and ponderable matter. Mathematical and
Physical Papers, 3, 484-515 (1890).
[2] O. Heaviside, The induction of currents in cores. The Electrician 12, 583-586 (1884).
[3] C. S. Whitehead, On the generalization of the functions ber x, bei x, ker x, kei x. Quart. J. Pure
Appl. Math. 42, 316-342 (1911).
[4] E. G. Richardson and E. Tyler, The transverse velocity gradient near the mouths of pipes in which
alternating or continuous flow of air is established. Proc. Phys. Soc. 42, 1-15 (1929).
[5] E. Reissner, Stresses and small displacements of shallow spherical shells. H. J. Mathematics and
Physics 25, 279-300 (1946), Correction, 27, 240 (1948).
[6] J. R. Womersley, Method for calculation of velocity, rate of flow and viscous drag in arteries when
the pressure gradient is known. J. Physiology 127, 553-563 (1955).
[7] G. N. Watson, A Treatise on the Theory of Bessel Functions, 2nd ed. Cambridge University Press,
Cambridge 1958.
[8] G. Petiau, La thkorie des fonctions de Bessel. Centre National de la Recherche Scientifique, Paris
1955.
[9] Y. Young and A. Kirk, Royal Society Mathematical Tables, vol. 10, Bessel Functions, Part IV,
Kelvin Functions. Cambridge University Press, Cambridge 1963.
[10] A. Abramowitz and I. E. Stegun, Handbook of Mathematical Functions. U.S. National Bureau of
Standards, Washington, DC 1965.
[11] N. W. McLachlan, Bessel Functions for Engineers, 2nd ed. The Clarendon Press, Oxford 1955.
[12] A. Apelblat and N. Kravitsky, Integral representation of derivatives and integrals with respect to the
order of the Bessel functions J,(t), Iv(t), the Anger function J,(t) and the integral Bessel function Ji~(t).
IMA J. Appl. Math. 34, 187-210 (1985).
[13] A. Apelblat, Derivatives and integrals with respect to the order of the Struve functions H~(t) and
Lv(t). J. Math. Anal. Appl. 137, 17-36 (1989),
[14] F. Oberhettinger and L. Badii, Tables of Laplace Transforms. Springer-Verlag, Berlin 1973.
Abstract
Integral representations of the Kelvin functions ber~ x and beiv x and their derivatives with respect
to the order are considered. Using the Laplace transform technique the derivatives are expressed in terms
of finite integrals. The Kelvin functions bern + 1/2 x and bein + 1/2 x can be presented in a closed form.
(Received: January 11, 1991; revised: April 18, 1991) | 3,269 | 9,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-30 | latest | en | 0.916496 |
https://probabilis.blogspot.com/2014/12/ | 1,643,122,274,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304835.96/warc/CC-MAIN-20220125130117-20220125160117-00285.warc.gz | 510,388,908 | 14,227 | ## Monday, December 29, 2014
### NFL Playoff Picks 2014-15
The NFL Playoffs start this weekend, and the top seeds (NE, DEN, SEA, and GB) are appropriate (representing 4 of the top 5 teams in Pyth (the lone missing team being Kansas City, who disappointingly didn't make the playoffs)), and these teams are heavily favored.
## Saturday, December 27, 2014
### Creating an NCAAF Model as a Function of Stadium Size
My friend roommate colleague (who's also my friend and roommate), "Stan Hooper", suggested that a decent model for predicting college football game outcomes would be rating teams based on the sizes of their stadiums.
I took the size of each team's stadium and predicted the winner of each game simply by which team had the larger capacity. I factored in home-field advantage by adding the average stadium capacity to the home team's stadium size (home-field advantage in terms of points is 3.5, which is approximately the average margin of victory by home teams (this season is 3.78)) in this equation.
On the season, the Home-Field Model predicted 60.96% of straight up winners. For comparison, my MDS Model predicted 71.71% correctly, and Vegas tabbed 75.11% correctly. Honestly I'm just relieved that a model created in 30 seconds by importing a Wikipedia page into a spreadsheet didn't beat the one I worked an entire semester on.
## Monday, December 22, 2014
### Kentucky is So Good They Broke My Model
The Matrix model uses a series of n x n matrices involving only wins and losses to calculate a "true" win percentage that factors in the strength of each team's opponents. There is a matrix for wins and a matrix for losses, which are then inverted, factoring in the total number of games each team has played, and then multiplied by an n x 1 vector that is each team's net wins (Note: n = number of teams (in NCAAB, 351)):
net wins = wins - losses
Each game simply is denoted by a "1". Therefore, the outputs of the win and loss matrices are bound between -1 and 1. If a team has a winning "adjusted" record, their rating is above 0; otherwise it's negative. I then standardize this rating to be analogous to win percentage, bound between 0 and 1:
i = initial Matrix rating; -1 <= i <= 1
i + 1 = i'
i' / 2 = f
f = final Matrix rating; 0 <= f <= 1
Kentucky has a rating of 1.005. As seen above, the final rating has an upper bound of 1. So the issue must be within the win and loss matrices themselves.
Kentucky's loss matrix rating is 1.000; this checks out, since they're undefeated and thus have 0 losses. The issue then is pinpointed to their win matrix: their rating is 1.010.
In their win matrix, they have 12 wins, and 12 total games (Note: only Division 1 games are included, but in this case all of Kentucky's opponents have been Division 1). This checks out too. Their strength of schedule is very high: 0.645 (0.500 is average). This is the only explanation I can offer as to why Kentucky's rating is so high: they've played a very tough slate of teams and beaten them all. Even so, it shouldn't be above 1.
A full rundown of the Matrix method can be found here on page 31, "Least Squares Ratings", written by Kenneth Massey. | 802 | 3,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-05 | latest | en | 0.963912 |
https://math.stackexchange.com/questions/2285965/how-to-find-the-vector-formula-for-the-bisector-of-given-two-vectors | 1,709,504,972,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476399.55/warc/CC-MAIN-20240303210414-20240304000414-00476.warc.gz | 389,366,813 | 36,331 | How to find the vector formula for the bisector of given two vectors?
There are two vectors called $\vec{a}$ and $\vec{b}$. Vector $\vec{c}$ is the bisector and it can be given as
$$\vec{c} = |b|\vec{a} + |a|\vec{b}$$
How to prove that? I have used the dot product method. But there I can't find the angle between them. Then I tried to do this by using unit vectors. Then I got an answer like in the given picture. The answer which I got. Can you help me to get the correct answer
• It appears that you are saying that $\mathbf{c}$ is a vector yet at the same time it is the sum of two magnitudes. But magnitudes are scalars, not vectors. Also your "equation" for $\mathbf{c}$ is incomprehensible as written and so is your diagram. May 18, 2017 at 4:50
• Remember that the angle between two vectors is defined as the value that makes $x \cdot y = |x||y|\cos{\theta}$ true, so try showing that the dot products of (a and c) and (b and c), when scaled appropriately, are the same thing. May 18, 2017 at 4:50
• John - I think there's just a bit of formatting issue. At a guess, I'd say the OP is trying to write $\vec{c} = |b|\vec{a} + |a|\vec{b}$, which is a perfectly fine vector expression. May 18, 2017 at 4:51
• @ConMan OK, I believe you are correct. That makes sense. May 18, 2017 at 4:52
• Can't you provide me an answer? May 18, 2017 at 5:01
Here's a purely geometric argument.
By definition, the sum of two vectors is equal to the diagonal of the parallelogram spanned by the vectors.
Now, observe that the two vectors $|b|\vec{a}$ and $|a|\vec{b}$ have exactly the same length. Therefore the parallelogram they span is a rhombus. The result then follows from the fact that the diagonal of a rhombus bisects its angles.
• [+1] I like this proof using figure geometry. So sad that that there are whole books with "geometry" in their title without any figure... May 18, 2017 at 5:17
• Very nice geometrical argument. May 18, 2017 at 5:32
Normalize $\vec{a}$ : $\frac{1}{||a||} \vec{a}$.
Normalize $\vec{b}$ : $\frac{1}{||b} \vec{b}$
Now add these two vectors to get:
$\vec{c}$ = $\frac{1}{||a||} \vec{a}$ + $\frac{1}{||b||} \vec{b}$.
Adding two unit vectors, vector addition, gives a resultant $\vec{c}$ that divides the angle between them.
$\vec{c}$ or any $\alpha \vec{c}$ are vectors that have this property.
Let $\alpha = ||a||$ $||b||$ to get:
$\vec{c} = ||b|| \vec{a}$ + $||a|| \vec{b}$.
Step 1 - normalise the original vectors. So define $\vec{\dot{a}} = \frac{\vec{a}}{|\vec{a}|}$ and similarly for $\vec{\dot{b}}$, then let $\vec{\dot{c}} = \vec{\dot{a}} + \vec{\dot{b}}$. It should be pretty simple to prove that the direction of $\vec{\dot{c}}$ is the same as the one of $\vec{c}$ in your post.
Step 2 - Find the angle between the new proposed bisector and the original vectors. So define $\alpha$ as the angle between $\vec{a}$ and $\vec{c}$, and then $\vec{\dot{a}} \cdot \vec{\dot{c}} = |\vec{\dot{a}}||\vec{\dot{c}}|\cos{\alpha} = |\vec{\dot{c}}|\cos{\alpha}$ since we set $|\vec{\dot{a}}| = 1$ in the first step. Similarly if $\beta$ is the angle between $\vec{b}$ and $\vec{c}$, then $\vec{\dot{b}} \cdot \vec{\dot{c}} = |\vec{\dot{c}}|\cos{\beta}$.
But, from the way they've been defined, $\vec{\dot{a}} \cdot \vec{\dot{c}} = \vec{\dot{a}} \cdot \vec{\dot{a}} + \vec{\dot{a}} \cdot \vec{\dot{b}} = |\vec{\dot{a}}| + \vec{\dot{a}} \cdot \vec{\dot{b}} = 1 + \vec{\dot{a}} \cdot \vec{\dot{b}}$, and you can show that the other dot product has the same value. So you can conclude that $\cos{\alpha} = \cos{\beta}$, and then all you have to do is show that the angles are in the same quadrant, and hence must be equal. | 1,171 | 3,654 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-10 | latest | en | 0.934761 |
https://math.stackexchange.com/questions/2034371/riemann-tensor-with-all-indices-lowered-on-a-2d-manifold | 1,713,937,905,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819067.85/warc/CC-MAIN-20240424045636-20240424075636-00744.warc.gz | 327,993,105 | 34,787 | # Riemann tensor with all indices lowered on a 2D manifold
I found the following in a textbook: On a 2-dimensional manifold the Riemann tensor with all indices lowered takes the form:
$R_{abcd} = R{g_{a}[_{c}g_{d}]_{b}}$
However I cannot see why this is true! I am sure it is correct but no amount of fiddling on my part gives the desired above result. Help would be much appreciated
• is $g$ the metric tensor? Nov 28, 2016 at 14:46
• Yes, I believe so. Nov 28, 2016 at 14:56
Since the Riemann tensor is anti-symmetric under exchange of the first two components, as well as under exchange of the last two components, we must have $a \neq b$ and $c \neq d$. Since $a,b,c,d$ can only take the values $1,2$ in two dimensions, this tells us that the only non-zero components of the Riemann tensor are $R_{1212}$ and permutations thereof.
If we work in Riemann normal coordinates at a given point, we obtain $$R_{abcd} = 2 R_{1212} g_{a[c} g_{d]b}$$ (just check all possible values of $a,b,c,d$ and use $g_{ab} = \delta_{ab}$ in normal coordinates). Taking the trace of both sides to compute the scalar curvature $R$ we obtain $$R = R_{abcd} g^{ac} g^{bd} = 2 R_{1212} g_{a[c} g_{d]b} g^{ac} g^{bd} = 2R_{1212}$$ Combining this with the result above we have $$R_{abcd} = R g_{a[c} g_{d]b}$$ and since this last equation is tensorial, it holds independent of the chosen coordinate system. | 425 | 1,388 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-18 | latest | en | 0.897947 |
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